Here's a small challenge:
Demonstrate the following without using conditional or indirect proof.
(Conditional proof makes it way too easy!)
1. (A ⊃ B)
2. (A ⊃ (B ⊃ C))
/ ∴ (A ⊃ C)
Rules you can use (see full post):
MP (p ⊃ q), p, / ∴ q
MT (p ⊃ q), ~q, / ∴ ~p
DS (p ∨ q), ~p, / ∴ q
HS (p ⊃ q), (q ⊃ r), / ∴ (p ⊃ r)
CD (p ⊃ q), (r ⊃ s), (p ∨ r), / ∴ (q ∨ s)
Add p, / ∴ (p ∨ q)
Simp (p · q), / ∴ p
Conj p, q, / ∴ (p · q)
Com (p · q) ⇔ (q · p), (p ∨ q) ⇔ (q ∨ p)
DM ~(p · q) ⇔ (~p ∨ ~q), ~(p ∨ q) ⇔ (~p · ~q)
DN p ⇔ ~~p
Assoc (p ∨ (q ∨ r)) ⇔ ((p ∨ q) ∨ r), (p · (q · r)) ⇔ ((p · q) · r)
Dist (p ∨ (q · r)) ⇔ ((p ∨ q) · (p ∨ r)), (p · (q ∨ r)) ⇔ ((p · q) ∨ (p · r))
Impl (p ⊃ q) ⇔ (~p ∨ q)
Contra (p ⊃ q) ⇔ (~q ∨ ~p)
Exp (p ⊃ (q ⊃ r)) ⇔ ((p · q) ⊃ r)
Taut (p ∨ p) ⇔ p, (p · p) ⇔ p
Equiv (p ≡ q) ⇔ ((p ⊃ q) · (q ⊃ p)), (p ≡ q) ⇔ ((p · q) ∨ (~p · ~q))
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4 comments:
Can we use premises such as (B ∨ ∼B)?
No, sorry. Being allowed to introduce tautologies would be nice; but unfortunately also would make it too easy.
I'm not trying to be sadisitc; I'm going by what my logic textbook allowed my students to do, such that I was forced to figure out how to do this without such tricks myself.
Besides, it makes more of a challenge of it.
I figured as much.
Here's my first attempt; I'm assuming the the second set of rules are things that you can swap in at any time, and not simply rules of inference with a single input. That's probably wrong, though it would seem that ((p ∨ q) ∨ r) ⇔ ((q ∨ p) ∨ r), which would otherwise be non-deducible from the current rules.
1. A ⊃ B
2. A ⊃ (B ⊃ C)
3. (A · B) ⊃ C (Exp 2)
4. (B · A) ⊃ C (Com 3)
5. B ⊃ (A ⊃ C) (Exp 4)
6. A ⊃ (A ⊃ C) (HS 1, 5)
7. (A · A) ⊃ C (Exp 6)
8. A ⊃ C (Taut 7)
Nothing wrong with your solution.
Good job!
Yes, the second set of rules (the last ten), unlike the first eight, can be used in either direction, and with any part of a line.
Hurely calls them "rules of replacement"; I call them "equivalence rules", but that can be confusing since one of them is *the* equivalence rule.
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