Single-Predicate Derivations

Transcription

1 Single-Predicate Derivations Let s do some derivations. Start with an easy one: Practice #1: Fb, Gb Ⱶ (ꓱx)(Fx Gx) Imagine that I have a frog named Bob. The above inference might go like this: Bob is friendly. Bob is green. Therefore, there exists something that is both friendly and green. That seems intuitive. Let s prove it: 1 (1) Fb A 2 (2) Gb A 1, 2 (3) Fb Gb 1, 2, I 1, 2 (4) (ꓱx)(Fx Gx) 3, ꓱI The derivation above simply requires conjunction-introduction followed by existentialintroduction. Let s do a more challenging one. Derive this sequent: Practice #2: (ꓯy)(Gy Lc), (ꓱx)Gx Ⱶ (ꓱz)Lz Remember that x, y, and z are variables, but c is a name. So, the argument might go like this: For all things, if that thing is God, then Chad is loved. God exists. Therefore, something exists that is loved. Here s the beginning of the derivation: Let s begin by assuming the quantifier-less portion of (1) for the purposes of ꓱE: 3 (3) Gx Ass. (ꓱE) * * Remember: The exception was only that our chosen variable (in this case x ) cannot occur freely in any prior assumption. And it doesn t! While x DOES occur in one prior assumption (on line 2), it is bound in line (2), not free. We can now use ꓯE to get rid of the universal quantifier in line (1). Don t forget that the y can be any variable. Below, I will make it an x : 1

2 3 (3) Gx Ass. (ꓱE) 1 (4) Gx Lc 1, ꓯE * 1, 3 (5) Lc 3, 4, E 1, 3 (6) (ꓱz)Lz 5, ꓱI * Remember: The exception was only that our chosen variable (in this case x ) cannot occur freely in the assumption that it rests on. And it doesn t! In fact, in line (1), x does not occur at all. We obtained the conclusion on line (6), but apparently we re not done yet. Why not? Reason: Notice that line (6) rests on (1) and (3). What we were aiming for in the conclusion on line (n) was that same formula, but resting instead on lines (1) and (2). What next? Don t forget that our assumption on line (3) was for the purpose of performing an ꓱE, but we haven t yet performed one! Let s do that: 3 (3) Gx Ass. (ꓱE) 1 (4) Gx Lc 1, ꓯE 1, 3 (5) Lc 3, 4, E 1, 3 (6) (ꓱz)Lz 5, ꓱI 2 (7) Gx (ꓱz)Lz 3, 6, I 1, 2 (8) (ꓱz)Lz 2, 7, ꓱE Remember that when we perform an ꓱE, it discharges the original assumption that was made for the purposes of the ꓱE i.e., it discharges line (3) while bringing back in the original quantified formula to rest on it instead i.e., the conclusion rests on (2). Let s do some more. Here are two equivalence relations between ꓯ and ꓱ that you will use frequently: Quantifier Negations (1) (ꓯx)Δx (ꓱx) Δx (2) (ꓱx)Δx (ꓯx) Δx Line (1) is intuitively true. For instance, Not all things are dogs means the same thing as At least one non-dog exists. (ꓯx)Dx (ꓱx) Dx Line (2) is also intuitive. For instance, Unicorns do not exist means the same thing as, All things are non-unicorns. (ꓱx)Ux (ꓯx) Ux 2

3 Remember when we learned that we could do without the operator? Well, we could also do without either the ꓯ or the ꓱ (we just need one of the two). For instance, we don t need ꓱ. Every time we wanted to write (ꓱx)Δx, we could just write (ꓯx) Δx instead. But, having both ꓯ and ꓱ makes logic a lot easier. To memorize these easily, just remember that, to convert from one quantifier to the other, you negate BOTH the quantifier AND the quantified wff. For instance, (ꓯx)Δx is equivalent to (ꓱx) Δx while (ꓱx)Δx (ꓯx) Δx. Let s do some derivations for these quantifier negations equivalence relations. Here s (2) from right to left: Practice #3: S124: (ꓯx) Lx Ⱶ (ꓱx)Lx Since the target is a negation, we ll try assuming it without the dash (for reductio): 1 (n) (ꓱx)Lx? Notice lines (1) and (2) contain Lx and Lx. These seem like good candidates for a contradiction. We can easily get Lx by using ꓯE on line (1). But, the only way we can get Lx is by assuming it (for ꓱE?). Let s see how far we can get with that: 3 (3) Lx Ass. (ꓱE) * 1 (4) Lx 1, ꓯE ** * This is permissible because x does not occur freely in any prior assumption. ** This is permissible because x does not occur freely in the assumption that it rests on namely, line (1). But, the contradiction we ve derived in line (5) does NOT rest on the assumption in (2). So, we cannot negate it. It looks like we can only use our contradiction to negate line (1). Let s do that: 3 (3) Lx Ass. (ꓱE) 1 (4) Lx 1, ꓯE 3 (6) (ꓯx) Lx 1, 5, I 1 (n) (ꓱx)Lx? 3

4 Note that we COULD combine (1) and (6) to get another contradiction at this point, but then that contradiction would rest on lines (1) and (3). Negating either of those lines won t help us. What we WANT is to negate our assumption in line (2)! If we could get a contradiction that rests on line (2), THEN we could negate it to get the conclusion. We can do that. Remember how ꓱE works. We assume the unbound wff for purposes of ꓱE, derive some other wff from that assumption, then use I, and finally use ꓱE to end up with a wff that rests ONLY on our original existentially quantified statement: 3 (3) Lx Ass. (ꓱE) 1 (4) Lx 1, ꓯE 3 (6) (ꓯx) Lx 1, 5, I - (7) Lx (ꓯx) Lx 3, 6, I 2 (8) (ꓯx) Lx 2, 7, ꓱE 1, 2 (9) (ꓯx) Lx (ꓯx) Lx 1, 8, I 1 (10) (ꓱx)Lx 2, 9, I Okay! That was a bit tricky. Practice #4: Let s do one more. This one is even trickier than the last. S125: (ꓯx)Lx Ⱶ (ꓱx) Lx As it turns out, we have to assume the negation of the conclusion again for purposes of a reductio. But, what then? 2 (2) (ꓱx) Lx Ass. (Red.) 1 (n) (ꓱx) Lx? Well, we ve assumed line (2) for the purposes of a reductio, so that means we need to get a contradiction somehow. It would be nice if we could derive (ꓯx)Lx since we already have its negation on line (1). So, let s aim for the following: 4

5 2 (2) (ꓱx) Lx Ass. (Red.) 1, 2 (n-2) (ꓯx)Lx (ꓯx)Lx (derived a contradiction to be used for reductio) 1 (n-1) (ꓱx) Lx 2, n-2, I (negating line (2) due to derived contradiction) 1 (n) (ꓱx) Lx n-1, E (simply getting rid of the double-negation) Ok, so, the target is (ꓯx)Lx which means our target is really just Lx since we can use ꓯI once we have that. So, let s assume the negation of our target for the purposes of ANOTHER reductio: 2 (2) (ꓱx) Lx Ass. (Red.) #1 3 (3) Lx Ass. (Red.) #2 (assume the opposite of target for reductio) 3 (4) (ꓱx) Lx 3, ꓱI (simply adding ꓱ to our assumption) 2, 3 (5) (ꓱx) Lx (ꓱx) Lx 2, 4, I (2+4 to derive a contradiction for reductio #2) 2 (6) Lx 3, 5, I (negating assumption from (3) due to contradiction) 2 (7) Lx 6, E (simply getting rid of the double-negation) 2 (8) (ꓯx)Lx 7, ꓯI * (simply adding ꓯ to line 7) 1, 2 (9) (ꓯx)Lx (ꓯx)Lx 1, 8, I (1+8 to derive a contradiction for reductio #1) 1 (10) (ꓱx) Lx 2, 9, I (negating assumption from (2) due to contradiction) 1 (11) (ꓱx) Lx 10, E (simply getting rid of the double-negation) Whew! We did it! * This is permissible because x does not occur freely in the assumption that it rests on namely, line (2). While x DOES occur in line (2), it is bound by the existential quantifier there. 5