Propositional Calculus and associated problems

Question

What I'm trying to prove: $P\to (Q\to S)$

Given Premises: $P\to (Q\to R)$ and $Q\to (R\to S)$

My approach:

$$Q\to (R\to S)$$ $$(Q\land R)\to S$$ $$(R\land Q)\to S$$ $$R\to (Q\to S)$$ also from the premises,

$$P\to (Q\to R)$$ $$(P\land Q)\to R$$ $$\dot{.\hspace{.075in}.}(P\land Q)\to (Q\to S) $$

I'm kinda stuck on what to do after this.

Should I do $\neg(P\land Q)\lor (\neg Q\lor S)$ ?

It seems like I'm beating around a bush.

It is one of the first things you prove about the Propositional Calculus. It would make this problem easy. — ancient mathematician, Sep 06 '20 at 15:02

score 2 · Accepted Answer · answered Sep 06 '20 at 14:47

2

Applying the axiom $[A \rightarrow (B \rightarrow C)] \rightarrow [(A \rightarrow B) \rightarrow (A \rightarrow C)]$ to your last line will give you $(P \wedge Q) \rightarrow S$.

answered Sep 06 '20 at 14:47

Qez

76
5

Is there an axiom like that? – Nimrod Sep 06 '20 at 14:50
There is in the system I use, any system you're using should have something equivalent to it at least. – Qez Sep 06 '20 at 14:52

Propositional Calculus and associated problems

1 Answers1