Introduction to Deductive Logic Summer 2004 87
Introduction Rule for ' ' (p. 157/173)
The ' ' Introduction rule allows us to write down a sentence of the form P Q whenever we have two previous and accessible subderivations, one starting with the assumption of P and ending with the sentence Q and the other starting with the assumption of Q and ending with the sentence P. It looks like this:
i Q Assumption
j P
n Q Assumption
m P
k P Q i–j, n–m I
The subderivations i–j and n–m can occur in either order. Introduction to Deductive Logic Summer 2004 88
Introduction and Elimination Rules for '~' (p. 152/168) The Introduction and Elimination rules for '~' are the most difficult to know how to apply. In all of the rules we have looked at so far, every SL-sentence that appeared in the rule-schema was a sentence that was either part of the input or the output of the inference rule. (Of course, many other sentences may be involved between the SL- sentences that explicitly appear in the schema.) This makes the rules relatively easy to use—your goal is in effect determined by the rule. The negation rules are not like that. The rules for adding or removing the '~' from some sentence P or ~P. involve the use of some sentence R that need bear no relation at all to P. This makes it harder to see what to aim for when you are using the negation rules. Introduction to Deductive Logic Summer 2004 89
Introduction Rule for '~' The '~' Introduction rule allows us to write down a sentence of the form ~P immediately after a subderivation which starts with an assumption of P and includes, under the scope of P, some sentence R and its negation ~R. It looks like this:
i P Assumption
j R
h ~R
k ~P i–h ~ I
The lines j and h can occur in either order. But one of the two must be immediately before line k and thus the last line of the subderivation. Introduction to Deductive Logic Summer 2004 90
Elimination Rule for '~' The '~' Elimination rule allows us to write down a sentence of the form P immediately after a subderivation which starts with an assumption of ~P and includes, under the scope of ~P, some sentence R and its negation ~R. It looks like this:
i ~P Assumption
j R
h ~R
k P i–h ~ E
The lines j and h can occur in either order. But one of the two must be immediately before line k and thus the last line of the subderivation. Introduction to Deductive Logic Summer 2004 91
Elimination Rule for ''
The '' Elimination rule is our most complicated rule. It allows us to write down a sentence of the form R if we have: 1) a disjunction P Q, 2) a subderivation which starts with an assumption of P and concludes with the sentence R, and 3) a subderivation which starts with an assumption of Q and concludes with the sentence R. It looks like this:
i P Q
j P Assumption
h R
m Q Assumption
n R
k R i, j–h, m–n I
The subderivations j–h and m–n can occur in either order. Introduction to Deductive Logic Summer 2004 92
Derivability in SD (p. 170/185)
A sentence P of SL is derivable in SD from a set of sentences of SL if and only if there is a derivation in SD in which all of the primary assumptions are members of and P occurs in the scope of only those assumptions.
├SD P =df P is derivable from in SD.
├SD P =df P is not derivable from in SD.
├SD P =df ├SD P. Introduction to Deductive Logic Summer 2004 93
Validity in SD (p. 171/186) An argument of SL is valid in SD if and only if the conclusion of the argument is derivable in SD from the set consisting of the premises. An argument of SL is invalid in SD if and only if it is not valid in SD.
Theorem in SD (p. 172/187)
A sentence P of SL is a theorem in SD if and only if P is derivable in SD from the empty set. (If an SL-sentence P is a theorem in SD, we also call it an SD-theorem.)
Thus, an SL-sentence P is an SD-theorem if and only if ├SD P. Introduction to Deductive Logic Summer 2004 94
Equivalence in SD (p. 173/187)
Sentences P and Q of SL are equivalent in SD if and only if Q is derivable in SD from { P } and P is derivable in SD from { Q }. (If SL- sentences P and Q are equivalent in SD, we also call them SD- equivalent.)
Thus SL-sentences P and Q are equivalent in SD if and only if
{P } ├SD Q and {Q } ├SD P.
Inconsistency in SD (p. 174/188)
A set of sentences of SL is inconsistent in SD if and only if both a sentence P of SL and its negation ~P are derivable in SD from
(i.e. iff for some SL-sentence P, both ├SD P and ├SD ~P ).
A set of sentences of SL is consistent in SD if and only if it is not inconsistent in SD. We also call a set of SL-sentences that is consistent (inconsistent) in SD SD-consistent (SD-inconsistent). Introduction to Deductive Logic Summer 2004 95
Soundness of a Logical System (p. 229/248) For some logical language L and for some logical system S based upon that language to say that S is sound is to say:
If some L-sentence P is derivable in S from a set of L-sentences , then entails P.
Put symbolically, for any L-sentence P and any set of L- sentences :
If ├S P then ╞L P. Or, more symbolically still, for any L-sentence P and any set of L-sentences :
├S P ╞L P.
Metatheorem 6.2/6.3.1 Soundness of SD
For any SL-sentence P and any set of SL-sentences ,
├SD P ╞SL P Introduction to Deductive Logic Summer 2004 96
Strong versus Weak Soundness Some presentations of logic distinguish between strong and weak soundness. In such presentations, a systems S is said to be weakly sound iff
├S P ╞L P. Whereas a system S is said to be strongly sound iff
├S P ╞L P. We won't be drawing upon this distinction in this course. (But it is worth noting in passing that there are two different notions here.) Introduction to Deductive Logic Summer 2004 97
Completeness of a Logical System (p. 236/256) For some logical language L and for some logical system S based upon that language to say that S is complete is to say:
If some L-sentence P is entailed by a set of L- sentences , then P is derivable from in S.
Put symbolically, for any L-sentence P and any set of L-sentences :
If ╞L P then ├S P Or, more symbolically still, for any L-sentence P and any set of L- sentences :
╞L P ├S P
Metatheorem 6.3/6.4.1 Completeness of SD
For any SL-sentence P and any set of SL-sentences ,
╞SL P ├SD P Introduction to Deductive Logic Summer 2004 98
Completeness of a Logical System Continued: There are also notions of strong and weak completeness in many presentations of logic. (These are defined in the same way as the parallel notions of strong and weak soundness.)
So, in such presentations, a systems S is said to be weakly complete iff
╞L P ├S P.
Whereas a system S is said to be strongly sound iff
╞L P ├S P.
We won't be drawing upon this distinction in this course, either. (But, once again, it is worth noting in passing that there are two different notions here.) Introduction to Deductive Logic Summer 2004 99
Linking Completeness and Soundness Once we have proved both the Completeness and Soundness of SD we have this result:
╞SL P ├SD P, i.e. a sentence P is truth-functionally entailed by a set iff there is an SD derivation of P from .
This in turn means that our syntactic system SD “hooks up with” our semantics for SL in just the way that we would like. In less vague terms, we will have shown that the entailment and derivability notions pick out exactly the same set of SL- sentences/SL-sentence pairs.
Thus, our derivability notion will coincide exactly with the entailment notion such that we can derive all and only the sentences that are entailed. Introduction to Deductive Logic Summer 2004 100
Decidability for SL (p. 247/267) We have in the truth-table method, a mechanical test for semantic properties of SL like truth-functional truth, truth-functional validity, truth-functional consistency, and truth-functional entailment.
Further, since we have1 both Completeness and Soundness for SD, we can use the truth-table method to test whether the corresponding SD-properties hold of sentences of SL. (E.g. we can test whether a sentence P is a theorem of SD and whether a sentence P is derivable in SD from a set of SL-sentences .)
1 “Have both Completeness and Soundness for SD” in the sense that they are provable (we shall sketch this shortly), though we have not yet shown them to hold. Introduction to Deductive Logic Summer 2004 101
Truth-Function (p. 220/239) A truth-function is a mapping, for some positive integer n, from each possible combination of n truth-values to a unique truth-value. i.e. If is an n-placed truth function and is an n-membered ordered set of truth-values, “feeding” to will yield a truth-value.
Example: The conjunction truth-function is characterized by the truth-table for “&”:
P Q P & Q
T T T T F F F T F F F F
For each 2-membered ordered set of truth-values (there are four such ordered sets) the conjunction truth-function maps that combination to a unique truth-value. Introduction to Deductive Logic Summer 2004 102
Metatheorem 6.2.1 Truth-Functional Completeness of SL (p. 222/241) Every truth-function can be represented by a sentence of SL which contains no sentential connectives other than ‘~’, ‘’, and ‘&’.
Characteristic truth-tables define truth-functions by exhaustively listing all arguments each function can take and mapping each combination to a unique truth-value.
There are four truth-functions of one argument:
T T T T T F T F F T F F F T F F
These pictures are like abbreviated truth-tables. We call them truth- function schema. Introduction to Deductive Logic Summer 2004 103
Metatheorem 6.2.1 Truth-Functional Completeness of SL Continued: There are 16 truth-functions of two arguments: the truth-functions characterized by the truth-tables for ‘&’, ‘’, ‘’, and ‘’, and 12 others:
T, T ñ F T, T ñ F T, T ñ F T, T ñ F T, F ñ F T, F ñ F T, F ñ F T, F ñ T F, T ñ F F, T ñ F F, T ñ T F, T ñ F F, F ñ F F, F ñ T F, F ñ F F, F ñ F
T, T ñ T T, T ñ T T, T ñ F T, T ñ F T, F ñ T T, F ñ F T, F ñ T T, F ñ T F, T ñ F F, T ñ T F, T ñ T F, T ñ F F, F ñ F F, F ñ F F, F ñ F F, F ñ T
T, T ñ F T, T ñ T T, T ñ F T, T ñ T T, F ñ F T, F ñ T T, F ñ T T, F ñ T F, T ñ T F, T ñ F F, T ñ T F, T ñ T F, F ñ T F, F ñ T F, F ñ T F, F ñ T Introduction to Deductive Logic Summer 2004 104
Metatheorem 6.2.1 Truth-Functional Completeness of SL Continued: n (2 ) Generally, there are 2 truth-functions of n arguments. (Why?)
So there are 256 truth-functions of three arguments, 65, 536 truth- functions of four arguments, 16, 777, 216 truth-functions of five arguments, and 18, 446, 744, 073, 709, 551, 616 truth-functions of 6 arguments.
Clearly we cannot hope to prove MT 6.2.1 by considering each case! Instead, we develop an algorithm for constructing a sentence of SL which represents any arbitrary truth-function. Introduction to Deductive Logic Summer 2004 105
Metatheorem 6.2.1 Truth-Functional Completeness of SL Continued:
Step 1: To represent an n-place truth-function we will need n atomic sentences of SL. We use the alphabetically first n atomic sentences of SL.
Step 2: We form for each row of the truth-table the characteristic sentence of that row. The characteristic sentence for a row is an iterated conjunction:
(. . . (P1 & P2) & . . . & Pn) We call the row we are constructing the characteristic sentence for the target row. The listed conjuncts of the characteristic sentence form a set of sentences, which we index by j.
The conjuncts for the characteristic sentence of the target row are constructed as follows: if the j-th value in the target row is T (if the value in the j-th column of that row is T) then Pj is the j-th atomic sentence. If the j-th value in that row is F then Pj is the negation of the j-th atomic sentence.
An example might help here. Suppose the target row is: T T F T F F Then the characteristic sentence of that row is: ((((A & B) & ~C) & D) & ~E) Introduction to Deductive Logic Summer 2004 106
Metatheorem 6.2.1 Truth-Functional Completeness of SL Continued:
Step 3 We identify the rows of the truth-function schema that have as output the value T. There are three possibilities to consider: 1) There are no such rows. 2) There is one such row. 3) There is more than one such row.
Step 3.1 If there are no such rows, we conjoin the characteristic sentence of the first row with its negation. The result will be a sentence that receives F on all truth-value assignments, and thus represents the truth-function. Introduction to Deductive Logic Summer 2004 107
Metatheorem 6.2.1 Truth-Functional Completeness of SL Continued:
Step 3.2 If there is one such row then the characteristic sentence of that row represents the truth-function.
Step 3.3 If there are more than one such rows we form the disjunction of the characteristic sentences of those rows. For example if rows 2, 5, and 9 (and no others) have as output the value T, we form the sentence:
((P2 P5) P9) where Pm is the characteristic sentence of the m-th row. This disjunction will represent the truth-function. Introduction to Deductive Logic Summer 2004 108
Metatheorem 6.2.1 Truth-Functional Completeness of SL Continued:
An example: Say we wish to represent the truth-function: T, T, T T T, T, F F T, F, T F T, F, F T F, T, T F F, T, F F F, F, T T F, F, F F
Then the characteristic sentences of the first, fourth, and seventh rows will be (we omit the other rows for speed): ((A & B) & C) ((A & ~B) & ~C), and ((~A & ~B) & C) We then form their disjunction: ((((A & B) & C) ((A & ~B) & ~C)) ((~A & ~B) & C)) Introduction to Deductive Logic Summer 2004 109
Metatheorem 6.2.1 Truth-Functional Completeness of SL Continued:
The disjunction that we formed was:
((((A & B) & C) ((A & ~B) & ~C)) ((~A & ~B) & C))
By construction, if we did a truth table for this sentence, it would receive T on truth-value assignments which either assign ‘A’, ‘B’ and ‘C’ all T, or ‘A’ T and ‘B’ and ‘C’ F, or ‘A’ and ‘B’ F and ‘C’ T, and it would receive F on all others.
Notice that this algorithm shows you how to construct an SL- sentences using only the sentential connectives ‘~’, ‘’, and ‘&’ to express any truth-function whatsoever.
This then is sufficient to prove MT 6.1.
Q.E.D. Introduction to Deductive Logic Summer 2004 110
Metatheorem 6.1a: Truth-Functional Completeness of SL with respect to ‘~’ and ‘’. (p. 225/245) We also can prove: Every truth-function can be represented by a sentence of SL which contains no sentential connectives other than ‘~’, and ‘’.
Notice that every conjunction P & Q of SL is truth-functionally equivalent to ~(~P ~Q). This means that we may transform the characteristic sentence yielded by our algorithm into a truth- functionally equivalent sentence employing only the connectives ‘~’, and ‘’. Q.E.D. Introduction to Deductive Logic Summer 2004 111
Truth-Functional Completeness of SL Continued:
Similarily, we could show that { ~, & } and { ~, } comprise truth- functionally complete sets of connectives. (How?) Are there any singleton sets of connectives that are truth-functionally complete? Yes, and no. No, as there are no connectives in our language SL that are complete. Yes, as there are binary truth-functions which we do not have as the meaning of any of our connectives but that are expressively adequate on their own. So, we can extend our language SL to a language SL+ by adding a single binary connective symbol for one of those truth-functions that are expressively adequate on their own. (See ex. 5 & 6 on p. 228/247.) Introduction to Deductive Logic Summer 2004 112
An Example of Mathematical Induction (p. 211/230) If we can show that 1) 0 has a particular property, and 2) If all positive integers less than or equal to k (where k is a non-negative integer) have that property, then k +1 has that property, we will have shown that all positive integers have that property. We prove that every SL-sentence has an equal number of left and right parentheses by a mathematical induction on the number of occurrences of sentential connectives in an SL-sentence.
(This example might seem boring and perhaps even obvious. It certainly is boring, but the point is to illustrate the technique with an example that is not itself of any independent interest or difficulty. Thus, the technique, rather than the example, will be on display.) Introduction to Deductive Logic Summer 2004 113
An Example of Mathematical Induction Continued: We need a numerical property that we can do an induction on to cover all sentences of SL. As we will often do, we pick the number of sentential connectives that occur in a given SL-sentence.
Basis: Every SL-sentence with 0 occurrences of sentential connectives is an atomic sentence and as such contains no parentheses left or right. It thus contains an equal number of left and right parentheses.
Inductive Step: If every SL-sentence with k or fewer occurrences of sentential connectives is such that the number of left and right parentheses are equal, then
every SL-sentence with k +1 occurrences of sentential connectives is such that the number of left and right parentheses are equal. Introduction to Deductive Logic Summer 2004 114
An Example of Mathematical Induction Continued: We need to show that on the assumption of the antecedent of the inductive step (call the antecedent the inductive hypothesis) that it follows that any SL-sentence P that has k +1 occurrences of sentential connectives has an equal number of left and right parentheses.
Since k is non-negative, k +1 is greater than or equal to 1, so P will have at least one occurrence of a sentential connective. So P will be of one of the forms ~Q, (Q & R ), (Q R ), (Q R ), (Q R ). We divide these into two cases:
Case 1:
P has the form ~Q. If ~Q contains k +1 occurrences of connectives, then Q contains k occurrences of connectives. But then, by the hypothesis of the induction, the number of left and right parentheses in Q is equal. But ~Q contains all and only the parentheses occurring in Q. So ~Q contains an equal number of left and right parentheses as well. Introduction to Deductive Logic Summer 2004 115
An Example of Mathematical Induction Continued: Case 2: P is of one of the forms (Q & R ), (Q R ), (Q R ), (Q R ). If P contains k +1 occurrences of sentential connectives, Q and R each contain at most k occurrences of sentential connectives. Thus, by the inductive hypotheses, they each contain an equal number of left and right parentheses. But P contains all of the parentheses of Q and of R, plus one additional left one, and one additional right one. And equals added to equals yield equals. So, P contains an equal number of left and right parentheses. So we have shown that on the assumption that all SL-sentences with k or fewer occurrences of sentential connectives have an equal number of right and left parentheses, all SL-sentences with k +1 occurrences of sentential connectives have an equal number of right and left parentheses. This, together with the basis of the induction shows that all SL- sentences have an equal number of right and left parentheses. Q.E.D. Introduction to Deductive Logic Summer 2004 116
Preliminaries for the Soundness Proof (p. 229/248)
Let and 1 be sets. If every member of is a member of 1 then we say that is a subset of 1 and that 1 is a superset of . Symbolically, Í 1. (The similarity to the SL-symbol for the conditional is not accidental. What might the connection be?) Example: Let be {1, 2, 3 } and be {1, 2, 3, 4, 854 }. Then is a subset of and is a superset of .
Lemma 6.3.1/6.3.2: (p. 230/249)
If ╞SL P, then for every superset n of , we have n╞SL P. Proof:
Assume ╞SL P and let n be any superset of . If every member of n is true, then every member of its subset is true.
Thus, because ╞SL P, P is also true. Therefore, n╞SL P. Introduction to Deductive Logic Summer 2004 117
Preliminaries for the Soundness Proof Continued: Lemma 6.3.2/6.3.3 (p. 232/249)
{ Q }╞SL R ╞SL Q R.
Lemma 6.3.3/6.3.4 (p. 233/249)
╞SL Q and ╞SL ~Q for some sentence Q is truth-functionally inconsistent.
Proofs: See exercise 3.6E 2b and 3b (p. 100/114) Introduction to Deductive Logic Summer 2004 118
Preliminaries for the Soundness Proof Continued: Lemma 6.3.4/6.3.5 (p. 233/249)
{ Q } is truth-functionally inconsistent ╞SL ~Q.
Proof: 1) Assume { Q } is truth-functionally inconsistent. 2) Then there is no truth-value assignment on which all of the members of { Q } are true.
3) Thus, if every member of is true on a particular truth-value assignment, Q is false on that assignment (if it were not, then { Q } would not be truth-functionally inconsistent, CTH [contrary to our hypothesis]). 4) But if Q is false on that assignment (one which makes all members of true), ~Q will be true on that assignment.
5) Thus ╞SL ~Q.
Q.E.D. Introduction to Deductive Logic Summer 2004 119
Outline of the Proof of Soundness of SD (MT 6.2/6.3.1) Recall that Soundness of SD says:
├SD P ╞SL P So, what we have to show is that if there is an SD-derivation of P from , then P is truth-functionally entailed by .
But recall that this means that there is no truth-value assignment which makes all of the members of true and P false. Thus, we need to show that the rules of SD are truth-preserving. (Why will this do?) We will use a mathematical induction to show that each sentence in a derivation is true if all of the undischarged assumptions in whose scope that sentence lies are true.
The basis clause will show that that the first sentence of a derivation is true if all of the assumptions in whose scope it lies are true. Introduction to Deductive Logic Summer 2004 120
Outline of the Proof of Soundness of SD Continued: The inductive step will show that: If [Hypothesis of the induction] the sentences on the first k lines of a derivation are true if all of the undischarged assumptions in whose scope those sentence lie are true then the sentence on the (k +1)-th line of a derivation is true if all of the undischarged assumptions in whose scope that sentence lies are true. The basis clause is fairly trivial. To prove the inductive step, we adopt the hypothesis of the induction and then show that each rule of SD is truth-preserving. Introduction to Deductive Logic Summer 2004 121
Outline of the Proof of Soundness of SD Continued: We present the structure of our argument symbolically. For any derivation, let
Pk be the sentence appearing on the k-th line of that derivation. Let k be the set of undischarged assumptions in whose scope Pk lies. We perform an induction on k:
Basis Clause: 1╞SL P1
Inductive Step: If i╞SL Pi for every positive integer i k, then k +1╞SL Pk +1.
Conclusion: For every positive integer k, k╞SL Pk. Introduction to Deductive Logic Summer 2004 122
Proof of the Soundness of SD:
Basis Clause:
To Show: 1╞SL P1 Proof:
1) P1 is the sentence appearing on the first line of a derivation. 2) Every derivation begins with an assumption (either a primary assumption or the assumption of a sub-derivation).
3) Thus, P1 is an undischarged assumption that lies within its own scope. That is to say that the set of undischarged
assumptions in whose scope P1 lies is { P1 }.
4) Because { P1 }╞SL P1, we conclude the basis clause is true. Introduction to Deductive Logic Summer 2004 123
Proof of the Soundness of SD Continued:
Inductive Step: Let k be an arbitrary positive integer and assume the inductive hypothesis (i.e. for every positive integer ik, i╞SL Pi ).
We must show that on this assumption it follows that k +1╞SL Pk +1.
We shall consider each way in which Pk +1 might be justified and show our thesis holds whichever justification is used.
(This is why Proof about Natural Deduction systems is more work than proof about axiomatic systems and why proof about the extended system SD+ would be more work than proof about SD.) We now consider particular cases. Introduction to Deductive Logic Summer 2004 124
Proof of the Soundness of SD Continued:
Case 1: Pk +1 is an Assumption.
1) Then Pk +1 is a member of k + 1 , the set of undischarged assumptions is whose scope Pk +1 lies.
2) Thus, if every member of k +1 is true, Pk +1 as a member of k +1 is true.
3) Thus k +1╞SL Pk +1.
k + 1 Pk +1 Assumption Introduction to Deductive Logic Summer 2004 125
Proof of the Soundness of SD Continued:
Case 2: Pk +1 is justified by Reiteration.
1) Then Pk +1 occurs earlier in the derivation as the sentence Pi appearing at some previous line i. 2) Every assumption that is undischarged at line i must still be
remain undischarged at line k+1, or else Pi would not be accessible for the application of Reiteration.
3) Therefore i is a subset of k +1 (i.e. every member of i is still an undischarged assumption at line k+1.)
4) But, by our hypothesis of induction, i╞SL Pi . Thus, since i is
a subset of k +1, it follows from L 6.3.1/6.3.2 that k +1╞SL Pi.
5) Since Pi is Pk +1, k +1╞SL Pk +1.
i Pi (= Pk+1 ) Justified some way or other as, by hypothesis, it appears as line i of a derivation. From now on, we omit all reference to the justification of the previous lines.
k + 1 Pk +1 i R Introduction to Deductive Logic Summer 2004 126
Proof of the Soundness of SD Continued:
Case 3: Pk +1 is justified by Conjunction Introduction.
1) Then Pk +1 is of the form Q & R. Q and R appear at previous lines of the derivation, call these lines j and h.
2) By the inductive hypothesis, h╞SL Q and j╞SL R.
3) h is a subset of k +1 and j is a subset of k +1. (If they were not both subsets of k +1, Q and R would not both be accessible at line k +1, contrary to our hypothesis.)
4) It thus follows from (2) and (3) by L 6.3.1/6.3.2 that k +1╞SL Q
and k +1╞SL R.
5) But whenever both Q and R are true, Pk +1 (which is Q & R ) is true.
6) Thus, k +1╞SL Pk+1.
j Q Lines j and h could be in either order. In the cases that follow we won't explicitly mention this when the order does not matter. h R
k + 1 Pk +1 = Q & R j, h &I Introduction to Deductive Logic Summer 2004 127
Proof of the Soundness of SD Continued:
Case 4: Pk +1 is justified by Conjunction Elimination. 1) Then some previous line, call it h, of the derivation is of the
form Q & Pk +1 or of the form Pk +1 & Q.
2) By the inductive hypothesis, h truth-functionally entails the conjunction at line h. 3) And whenever a conjunction is true, both of its conjuncts are
true. Thus, h╞SL Pk+1. 4) Since all assumptions that have not been discharged at line h remain undischarged at line k +1 (else the conjunction
would not be accessible at line k +1), h is a subset of k+1.
5) It follows from (3) and (4) by L 6.3.1/6.3.2 that k +1╞SL Pk+1.
h Q & Pk +1 or Pk +1 & Q
k + 1 Pk +1 h &E Introduction to Deductive Logic Summer 2004 128
Proof of the Soundness of SD Continued:
Case 5: Pk +1 is justified by Disjunction Introduction.
1) Then Pk +1 is of the form Q R and some previous line of the derivation, call it h, is of the form Q or of the form R.
2) By the inductive hypothesis, h truth-functionally entails the sentence at line h. The sentence at line h is one of the
disjuncts of Pk +1 (i.e. of Q R ), so whenever it is true, so is Pk +1.
3) Thus h╞SL Pk+1. 4) Since all assumptions that have not been discharged at line h remain undischarged at line k +1 (else the sentence at line
h would not be accessible at line k+1), h is a subset of k+1.
5) So from (3), (4) and L 6.3.1/6.3.2 it follows that k +1╞SL Pk+1.
h Ph (Ph = Q or Ph = R)
k + 1 Pk +1 ( Pk + 1 = Q R) h I Introduction to Deductive Logic Summer 2004 129
Proof of the Soundness of SD Continued:
Case 6: Pk +1 is justified by Conditional Elimination. 1) Then there are two previous lines of the derivation at which
appear the sentences Q and Q Pk +1. Call these lines j and h respectively.
2) By the inductive hypothesis, h╞SL Q Pk +1 and j╞SL Q. 3) Since the sentences at lines h and j are accessible at line
k+1 by hypothesis, both j and h are subsets of k +1.
4) Thus k+1╞SL Q Pk +1 and k+1╞SL Q . (From (2) and (3) by L 6.3.1/6.3.2.)
5) But since Pk +1 is true if both Q Pk +1 and Q are true,
k +1╞SL Pk+1.
j Q
h Q Pk+1
k + 1 Pk +1 j, h E Introduction to Deductive Logic Summer 2004 130
Proof of the Soundness of SD Continued:
Case 7: Pk +1 is justified by Biconditional Elimination. 1) Then there are two previous lines of the derivation, line h at which appear the sentence Q and line j at which appears the
sentence Pj (which is either Q Pk +1 or Pk +1 Q ).
2) By the inductive hypothesis, h╞SLQ and j╞SL Pj.
3) Since lines h and j are accessible at line k +1, h and j are subsets of k +1.
4) Thus, by L 6.3.1/6.3.2 from (2) and (3) k +1 truth-functionally entails both Q and the biconditional appearing on line j.
5) But Pk+1 is true whenever both Q. and the biconditional appearing on line j are true.
6) Thus, k +1╞SL Pk+1.
j Pj ( Pj = Q Pk +1 or Pj = Pk +1 Q )
h Q
k + 1 Pk +1 j, h E Introduction to Deductive Logic Summer 2004 131
Proof of the Soundness of SD Continued:
Case 8: Pk +1 is justified by Conditional Introduction.
1) Then Pk +1 is of the form Q R where Q is an assumption for a subderivation commencing at line h and R appears at line j under the scope of the assumption at line h.
2) By the inductive hypothesis, j╞SL R. 3) Because the subderivation in which R is derived from Q is accessible at line k+1, every assumption that is undischarged at line j is undischarged at line k+1, save the assumption Q that begins the subderivation.
4) So j is a subset of k+1 { Q }.
5) Since j╞SL R, it follows by L 6.3.1/6.3.2 that k+1 { Q }╞SL R.
6) Then, by L 6.3.2/6.3.3 it follows that k+1╞SL Q R, and thus
that k+1╞SL Pk +1.
h Q Assumption
j R
k + 1 Pk +1 = Q R h–j I Introduction to Deductive Logic Summer 2004 132
Proof of the Soundness of SD Continued:
Case 9: Pk +1 is justified by Negation Introduction.
1) Then Pk +1 is of the form ~Q and there are three previous lines of the derivation where Q is an assumption for a subderivation commencing at line h, R appears at line j under the scope of the assumption at line h and ~R appears at line m under the scope of the assumption at line h.
2) By the inductive hypothesis, j╞SL R and m╞SL~R. 3) Because the subderivation deriving lines j and m from the assumption at line h is accessible at line k+1, every assumption that is undischarged at line j is undischarged at line k+1, save the assumption of Q at line h which commences the subderivation. The same is true of the assumptions that are undischarged at line m.
4) Thus, j and m are subsets of k+1 { Q }.
5) Thus, from (2) and (4) by L 6.3.1/6.3.2 k+1 { Q }╞SL R and
k+1 { Q }╞SL ~R. Introduction to Deductive Logic Summer 2004 133
Proof of the Soundness of SD Continued:
Case 9: Pk +1 is justified by Negation Introduction Continued:
6) By L 6.3.3/6.3.4 we conclude that k+1 { Q } is truth- functionally inconsistent.
7) Thus, from L 6.3.4/6.3.5 it follows that k+1╞SL ~Q, and thus
that k+1╞SL Pk +1.
h ~Q Assumption
j R m ~R
k + 1 Pk +1 = Q h–m ~ E
Case 10: Pk +1 is justified by Negation Elimination. See exercise 6.3.3 Introduction to Deductive Logic Summer 2004 134
Proof of the Soundness of SD Continued:
Case 11: Pk +1 is justified by Disjunction Elimination. 1) Then there are five previous lines of the derivation:
i.) line h which is of the form Q R.
ii.) line j which is the assumption of Q for a subderivation.
iii.) line m, which is Pk +1 in the scope of the assumption of Q at line j.
iv.) line n which is the assumption of R for a subderivation.
v.) line p, which is Pk +1 in the scope of the assumption of Q at line n.
h Q R
j Q Assumption
m Pk +1
n R Assumption
p Pk +1 k + 1 Pk +1 h, j–m, n–p E Introduction to Deductive Logic Summer 2004 135
Proof of the Soundness of SD Continued:
Case 11: Pk +1 is justified by Disjunction Elimination Continued:
2) By the inductive hypothesis, h╞SL Q R, m╞SL Pk +1, and
p╞SL Pk +1. 3) Because the two subderivations are accessible at line k+1,
the set of undischarged assumptions m is subset of
k+1 { Q }, and the set of undischarged assumptions p is
a subset of k+1 { R }.
4) Then by L 6.3.1/6.3.2, k+1 { Q }╞SL Pk +1 and
k+1 { R }╞SL Pk +1.
5) Further, since Q R at line h is accessible at line k+1, h is a subset of k +1.
6) Thus, since h╞SL Q R, it follows by L 6.3.1/6.3.2 that
k+1╞SL Q R. 7) Now consider any truth-value assignment on which every
member of k+1 is true. Since k+1╞SL Q R, Q R is also true on that assignment. Thus either Q or R is true on that assignment. Introduction to Deductive Logic Summer 2004 136
Proof of the Soundness of SD Continued:
Case 11: Pk +1 is justified by Disjunction Elimination Continued: 8) If Q is true on that assignment, then every member of
k+1 { Q } is true on that assignment and thus since
k+1 { Q }╞SL Pk+1, Pk+1 is also true on that assignment. 9) Similarly if R is true on that assignment, then every member
of k+1 { R } is true on that assignment and thus since
k+1 { R }╞SL Pk+1, Pk+1 is also true on that assignment. 10) Either way (whether Q or R is true on that assignment) it
follows that Pk+1 is true on any truth-value assignment on
which every member of k+1 is true.
11) Thus k+1╞SLPk +1. Introduction to Deductive Logic Summer 2004 137
Proof of the Soundness of SD Continued:
Case 12: Pk +1 is justified by Biconditional Introduction.
1) Then Pk +1 is of the form Q R there are four previous lines of the derivation:
i.) line h which is the assumption Q for a subderivation.
ii.) line j, which is R and is in the scope of the assumption at line h.
iii.) line m which is the assumption R for a subderivation.
iv.) line n, which is Q and is in the scope of the assumption at line m.
h Q Assumption
j R
m R Assumption
n Q
k + 1 Pk +1 (= Q R ) h–j, m–n I Introduction to Deductive Logic Summer 2004 138
Proof of the Soundness of SD Continued:
Case 12: Pk +1 is justified by Biconditional Introduction Continued: 3) Because the two subderivations h-j and m-n are accessible at line k+1 (else that line could not be justified by appeal to
them), j is a subset of k+1 { Q } and n is a subset of
k+1 { R }.
4) Then, by L 6.3.1 k+1 { Q }╞SL R and k+1 { R }╞SL Q. 5) Now consider any truth-value assignment on which each
member of k+1 is true. If R is true on that assignment then
so is Q because k+1 { R }╞SL Q. If R is false on that
assignment then so is Q because k+1 { Q }╞SL R. 6) Either way Q and R have the same truth-value, and thus Q R is true on every truth-value assignment on which
every member of k+1 is true.
7) Thus k+1╞SL Pk +1. Introduction to Deductive Logic Summer 2004 139
Proof of the Soundness of SD Continued: This completes out proof of the inductive step.
We assumed that i╞SL Pi for all positive integers i k and, on that assumption, showed that k+1╞SL Pk +1. We did this by considering each possible way the sentence appearing on line k+1 might be justified.
Having shown that on our assumption, no matter how Pk +1 is justified, k+1╞SL Pk +1 we have shown:
If the sentences on the first k lines of a derivation are true if all of the undischarged assumptions in whose scope those sentence lie are true then the sentence on the (k + 1)th line of a derivation is true if all of the assumptions in whose scope that sentence lies are true. This, together with the basis, shows us that all sentences of SL appearing in any derivation are such that they are truth-functionally entailed by the set of undischarged assumptions in whose scope they lie. Introduction to Deductive Logic Summer 2004 140
Proof of the Soundness of SD Continued:
In particular, we have shown that the last line of any derivation of P is such that it is truth-functionally entailed by the set of undischarged assumptions in whose scope it lies. But the undischarged assumptions at the last line of a derivation of P are the primary assumptions of the derivation. Thus, we have shown:
├SD P ╞SL P
Q.E.D. [Finally!]
Some Corollaries:
├SD P ╞SL P
Let be an argument. Then we have the following: is valid in SD is truth-functionally valid.