Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27P Q ¬Q ¬PP Q ¬P ᴠ QThe Law of Contraposition; the second conditional in this equivalence is called the contrapositive of the first. Note that this equivalence mirrors the inference pattern modus tollens. The first disjunct here covers the cases where the antecedent of the conditional is false (¬P) and thus doesn’t trigger the conditional relation; the second disjunct covers the one other case where the conditional is true, namely, when the condition is triggered by a true antecedent and thus yields a true consequent (Q).Equivalences involving conditionals¬(P Q) P ᴧ ¬QP ↔ Q (P Q) ᴧ (Q P)P ↔ Q (P ᴧ Q) ᴠ (¬P ᴧ ¬Q)Note that this equivalence reflects the second row of the truth table for the material conditional (where the conditional is false).This equivalence reflects one way of understanding the biconditional: as a two-way conditional relationship.This equivalence reflects another way of understanding the biconditional: as the covariance of truth values.Equivalences involving conditionals (continued)Conditional Elimination: Elim aka modus ponensIf you know P Q and also P individually, then you can assert Q individually as well (because P Q instructs you that Q follows from P).1 P Q2 P3 Q Elim: 1, 2Conditional Introduction: Intro(aka ‘Conditional Proof’)If you can prove within a subproof that premising P leads to the conclusion Q, then you can close the subproof and assert P Q one level back, citing the entire subproof.1 P 2 Q3 P Q Intro: 1-2P QP QNote that the only way to prove a conditional statement (e.g., when the goal of your proof is a conditional statement such as P Q) is to open a subproof of the sort we just saw that premises the antecedent of the conditional statement you want to prove and concludes with the consequent. So, anytime the goal of your proof is a conditional statement, you can go ahead and sketch in this overall structure, then worry about how to get from P to Q.Biconditional Elimination: ↔ Elim If you have a biconditional involving P and Q (i.e., P ↔ Q or Q ↔ P), and you also have either P or Q individually, then you can conclude the other (P or Q) individually. For example:1 P ↔ Q2 Q3 P ↔ Elim: 1, 2Biconditional Introduction: ↔ IntroThis is just like Conditional Introduction, except that you need two subproofs instead of just one, in order to prove that the conditional relationship works two ways:1 P This first subproof proves the P Q relation2 Q3 Q This second subproof proves the Q P relation4 P5 P ↔ Q ↔ Intro: 1-2, 3-4 The biconditional conclusion merges both relationsAs with material conditionals, the only way to prove a biconditional statement (e.g., when the goal of your proof is a biconditional such as P ↔ Q) is to open two subproofs which together prove both directions of the biconditional relation.Proof 8.29 (P Q) ↔ (¬P ᴠ Q)Here is proof 8.29 from your homework. Let’s work it step-by-step.Notice that this argument contains no premise, which means the conclusion is a tautology (recall that any valid argument—and this one is indeed valid—whose conclusion is provable without any premises must have a tautology as its conclusion). Because there is no premise, you’ll have to take your cue for how to tackle this proof entirely from the conclusion. Fortunately, the fact that the conclusion is a biconditional tells us how you’ll have to proceed here: you need to open two subproofs, with each subproof ultimately proving one of the two ‘directions’ of the biconditional relation in the conclusion.1 Proof 8.292 P Q ¬P ᴠ Q ¬P ᴠ Q P Q(P Q) ↔ (¬P ᴠ Q) ↔ Intro:Here are the two subproofs sketched out in skeleton form. Notice that the first one takes as its premise the first half of the biconditional relation found in the conclusion. This first subproof will attempt to show that the second half of the biconditional logically follows from that premise. The second subproof does just the opposite, seeking to prove that the second half of the biconditional logically leads to the first half of the biconditional.Notice too that you can go ahead and put down below the rule that will ultimately justify the conclusion of the proof (↔ Intro). It’s wise to go ahead and write this down now so that you won’t later forget what your overall strategy was. You can fill in the citation steps for the rule later once you know all the steps of the proof.Now, you need to focus on the first subproof. How will you get from the premise (P Q) to the conclusion of this subproof (¬P ᴠ Q)? The premise doesn’t really offer any cues for a strategy … and the disjunction in the conclusion isn’t really any help either (you can’t base a disjunction elim strategy off it—you would instead need a disjunction in the premise or higher up in the proof somewhere in order to pursue that strategy). But there is nothing else up there. Hmmm ….1 Proof 8.292 P Q3 ¬(¬P ᴠ Q) ¬P ᴠ Q ¬ Intro: ¬P ᴠ Q P Q(P Q) ↔ (¬P ᴠ Q) ↔ Intro:Luckily, at the last minute you remember that when all else fails, the negation intro strategy can often come to the rescue, as it does here. Just open a new sub-subproof and premise the opposite of the goal you need to reach. So, make your premise ¬(¬P ᴠ Q), which is the negation of the first subproof goal ¬P ᴠ Q. If you can show that this premise leads to a contradiction, you’ll be able to assert the goal on the basis of ‘proof by contradiction’ (i.e., negation intro). Now your challenge has been refined: How can you get from the new sub-subproof premise ¬(¬P ᴠ Q) to a contradiction?This is the trickiest part of the entire proof. You can’t reach the new sub-sub-goal (i.e., contradiction) directly on the basis of our wonder-strategy, negation intro, because there’s no way to premise the opposite of a contradiction (i.e., you can’t negate the contradiction symbol). But, perhaps you could think of an intermediate goal that you could reach by negation intro, and then move from that intermediate goal to the contradiction . . .1 Proof 8.292 P Q3 ¬(¬P ᴠ Q)4 ¬P P ¬ Intro:
View Full Document
Unlocking...