# Propositional logic, valid entailment

I have an old exam-question but don't know how to solve this type of problems. So If someone could give me a hint it would be much appreciated.

Is the following entailment valid?

$$(p \rightarrow q) \rightarrow r, \neg r\land \neg s, (q \rightarrow p) \lor t, t \rightarrow (r \lor p) \vdash t$$

If I assume that the right handside is false$$(\neg t)$$ and show that the left handside holds, then the entailment is not valid. But how?

EDIT

For $$\neg r \land \neg s$$ to be true then both  r and s must be false. And if r is false then $$(p \rightarrow q)$$ must be false.  And since $$(q \rightarrow p)$$ must be true, then p is true and q is false.

When t is negated, all premises is satisfiable and therefor the entailment is not valid. Am I right?

• Have you made any progress? – Graham Kemp Aug 21 '18 at 2:14
• Oh. Thanks for asking. Not really. I'm I a making it more complicated when adding a negation to 't'(right hand side)? – Lars Logik Oct 14 '18 at 16:23
• Yes, indeed; since the conclusion may be false with all premises satisfied, therefore $t$ is not entailed by those premises. – Graham Kemp Oct 14 '18 at 21:14
• I have now tried this when assuming t is false and when t is true. Both times all premises is true if p is true and q is false. Is it then correct to say that the entailment is not valid but satisfiable? – Lars Logik Oct 16 '18 at 11:52
• Yes, and therefore.... – Graham Kemp Oct 16 '18 at 11:56

$$\{(p \rightarrow q) \rightarrow r, \neg r\land \neg s, (q \rightarrow p) \lor t, t \rightarrow (r \lor p), \neg t\}$$
Clearly, the negation of the conclusion may be satisfied when $t$ is false. That also satisfies the fourth premise, but the third premise would then only be satisfied when $q$ implies $p$.
$$\{(p \rightarrow q) \rightarrow r, \neg r\land \neg s, (q \rightarrow p) \lor \bot, \bot \rightarrow (r \lor p), \neg \bot\}$$
• Yes, @LarsLogik . $t\to(r\vee p)$ is satisfied when $\lnot t$ is; so that does not fix $r$ or $p$. For that, look to the second premise, $\lnot r\land\lnot s$. – Graham Kemp Oct 14 '18 at 21:07