Natural Deduction in P 7 2 Derivations Strategies

Natural Deduction in P 7. 2 Derivations: Strategies and Notes

Proofs What to Do—and what NOT to do

The Friendly Rules • ∀E and ∃I are unrestricted and do not use subderivations. • Cautions - ∀E: All instances of quantified variable must be replaced by same constant. E. g. you can’t get Rab from (∀x)Rxx - ∃I: Different constants cannot be replaced with the same variable. E. g. from Gfk you can’t get (∃x)Gxx. Or, on UD: ℕ from 5 > 4 get (∃x)x > x!

P. 253 Example 1 Fe P 2 (∀z)(Fz → Gz) P 3 Fe → Ge 2 ∀E 4 Ge 1, 3 → E 5 Fe ∧ Ge 1, 4 ∧I 6 (∃x)Fx ∧ Gx) 5 ∃I ⊢ (∃x)(Fx ∧ Gx) • We want an existentially quantified conjunction: the ‘Fx’ bit will come from 1. • Get rid of the quantifier in 2 so that we can apply SDE rules. • Now stick on the quantifier and replace constants with variable. DONE!

The Not-So-Friendly Rules • Restrictions for ∀I and ∃E: (i) �� does not occur in an undischarged assumptions (ii) �� does not occur in (∀x)ℙ (for ∀I) or (∃x)ℙ (for ∃E) • Restriction for ∃E: in addition to (i) and (ii) above (iii) �� does not occur in ℚ

P. 252 ∀I Example 1 (∀x)Fx P 2 (∀z)(Fz → Gz) P 3 Fe 1 ∀E 4 Fe → Ge 2 ∀E 5 Ge 3, 4 →E 6 Fe ∧ Ge 3, 5 ∧I 7 (∀x)(Fx ∧ Gx) ⊢ (∀x)(Fx ∧ Gx) 6 ∀I • ∀E is unrestricted so we can straightaway replace (∀x)Fx with Fe and (∀z)(Fz → Gz) with Fe → Ge. Any constant will do. • Since e was arbitrarily chosen—says in 1 that anything we could have chosen is F—we can infer that everything is F.

P. 253 ∃E Example: What Not to Do 1 (∃x)Fx P 2 (∀z)(Fz → Gz) P 3 Fe 1 ∃E 4 Fe → Ge 1, 3 → E 5 Ge 3, 4 → 6 Fe ∧ Ge 3, 5 ∧I 7 (∃x)(Fx ∧ Gx) ⊢ (∃x)(Fx ∧ Gx) MISTAKE! 6 ∃I • You can’t just go from (∃x)Fx—something is F—to this thing e is F • You have to enter a World of What-IF subderivation: ‘something or other is F (says 1)—let’s call whatever it is ‘e’ and see what happens…

P. 253 ∃E Example: What to Do 1 (∃x)Fx P 2 (∀z)(Fz → Gz) P ⊢ (∃x)(Fx ∧ Gx) 3 Fe A 4 Fe → Ge 2 ∃E 5 Ge 3, 4 → E 6 Fe ∧ Ge 3, 5 ∧I 7 (∃x)(Fx ∧ Gx) 8 (∃x)(Fx ∧ Gx) 6 ∃I 1, 3 – 7 ∃E • ∃E rule begins with assumption: ‘Some thing(s) is F. Let’s choose an arbitrary object from those that are F and call it “e”—and see what happens. ’ • We get ‘(∃x)(Fx ∧ Gx)’ from that safe assumption so it really is so!

Choice of Constants • Observe the restrictions on choice of constants for the ∀I and ∃E rules! • Where there are no restrictions, choose constants that will allow you to connect to other sentences in the proof. - E. g. if you have ‘(∀x)(Fx→ Gx)’ and ‘Fa’, and want to get ‘Ga’ then obviously applying ∀E to get ‘Fb → Gb’ is legal but stupid! • Don’t paint yourself into a corner! - Look ahead: if the legal choice of a constant makes it impossible to apply one of restrictive rules later on try using a different constant!

To paint oneself into a corner: to get oneself into a difficulty from which one cannot extricate oneself

P. 255: On Being Painted Into Corner 1 (∀x)(∀y)Cxy P 2 (∀y)Cby 1 ∀E 3 Cbb 2 ∀E 4 (∀y)Cyb 3 ∀I 5 (∃x)(∀y)Cxy 4 ∃I ⊢ (∃x)(∀y)Cyx MISTAKE! • We need to get rid of quantifiers and put in quantifiers, flipping things around. • ‘b’ occurs in ‘(∀y)Cyb’ in line 2 so line 4 is illegal according to ∀I rule restriction (ii) • Rationale for restriction: Cbb (line 3) says that b Cs itself! From that we can’t infer that everything Cs b! (I’m identical to myself—it is not the case that everyone is identical to me!)

P. 255: On Not Being Painted Into Corner 1 (∀x)(∀y)Cxy P 2 (∀y)Cby 1 ∀E 3 Cbc 2 ∀E 4 (∀y)Cyc 3 ∀I 5 (∃x)(∀y)Cxy 4 ∃I ⊢ (∃x)(∀y)Cyx • This time we looked ahead and saw that even though it was legal to get ’Cbb’ in line 3 it would get us into trouble because we had to apply the ∀I rule later on. • We picked ‘c’ as the constant for line 3 instead of ‘b’ so proof went fine!

P. 256: On Being Painted Into Corner 1 (∀y)Bay P 2 Baa 1 ∀E 3 (∃x)Bxa 2 ∃I 4 (∀y)(∃x)Bxy 3 ∀I ⊢ (∀y)(∃x)Bxy MISTAKE! • Same problem as before: ‘a’ appears in ‘(∀y)Bay, so we’ll end up violating restriction (ii) on ∀I later in the proof. • Rationale for restriction: 3 says there’s something that Bs this particular thing a, which we were talking about in 1… • From which we can’t infer 4—which says there’s something that Bs everything.

P. 256: On Not Being. Painted Into Corner 1 (∀y)Bay P 2 Bab 1 ∀E 3 (∃x)Bxb 2 ∃I 4 (∀y)(∃x)Bxy 3 ∀I ⊢ (∀y)(∃x)Bxy • Same fix as before: pick a different constant! • Now 3 just says there’s something that Bs any arbitrarily selected object which, for convenience, we’ve called ‘b’. • Since b is arbitrarily chosen it’s any-old-thing so we can in 4 which here say that there’s something, x, that Bs any-old-thing.

Thu Oct 29 Exercises More proofs! Try them! I’ll work them in the Zoom recording.

Proofs Using ∀E and ∀I

Proofs Using ∀E and ∀I

Proof Using ∃I and ∃I

A Proof That Uses All the Quantifier Rules!

It’s Complicated…Or Not Complicated