0

Currently stuck on looking at some inferences to identify the basic inference rule of which it is a substitution instance. Examples and questions taken from book "Fuzzy set theory" by Klir/St Clair/Bo Yuan (which frustratingly does not contain the solutions and I cannot find them anywhere on internet).

Example which I understand

Consider the following premises given as true

(P1) p v q (P2) ¬r → ¬p (P3) r → s (P4) ¬s

(c) ∴ q (conclusion to be established)

Proving conclusion c using basic inference forms

(r1) ¬r (P3), (P4), MT

(r2) ¬p (P2), (r1) MP

(r3) q (P1, (r2) DS

where r3 is the proved conclusion, using basic inference rules MT (Modus Tollens), MP (Modus Ponens) and DS (Disjunctive Syllogism).

Stuck getting started on proving the following:

(p1) (p^¬q)→(r v s)

(p2) p^¬q

(c) ∴ r v s

(P1) (¬p→q) ^ (r v ¬s)

(c) ∴ [(¬p→q) ^ (r v ¬s)] v (¬p→s)

(P1) [(p↔¬q) ^ s] ^ (¬p↔q)

(c) ∴(p≡¬q) ^ s

Much obliged for any support, working hard to get my head around this. Thanks!

Grumpybeard
  • 1
  • The first one needs one application only of Modus Ponens and it's done. – Mauro ALLEGRANZA Oct 22 '19 at 11:04
  • For the second one you need Disjunction introduction (aka : Addition). – Mauro ALLEGRANZA Oct 22 '19 at 11:05
  • So proof for first as follows? (r1) p^¬q (p2) (Modus Ponens)

    (c) ∴ r v s

     – Grumpybeard Oct 22 '19 at 11:20
  • Thanks but really not clear on how I establish the conclusion [(¬p→q) ^ (r v ¬s)] v (¬p→s) for the second one. I can see the addition of (¬p→s) but dont understand how "if not p then s) is derived. – Grumpybeard Oct 22 '19 at 11:33
  • Not clear what is not clear… The rule of inference called Addition allows us to derive from formula $\alpha$ the disjunction $\alpha \lor \beta$, for $\beta$ whatever. Thus, you have only to apply it to $(¬p → q) \land (r \lor ¬s)$ as $\alpha$ and $(¬p → s)$ as $\beta$. – Mauro ALLEGRANZA Oct 22 '19 at 11:46
  • Ok the "for β whatever" clarified, thanks. Any idea on third - Simplification? – Grumpybeard Oct 22 '19 at 12:04

0 Answers0