# Prove that $\{ \phi \rightarrow(\psi \rightarrow \theta)\} \vdash \phi \wedge \psi \rightarrow \theta$?

Please, can you check is my solution of this problem $$\{ \phi \rightarrow(\psi \rightarrow \theta)\} \vdash \phi \wedge \psi \rightarrow \theta$$ good?

First, I rewrote it like $$\{ \phi \rightarrow(\psi \rightarrow \theta), \phi \wedge \psi \} \vdash \theta$$. After that

1) $$\phi \rightarrow(\psi \rightarrow \theta)$$ - premise

2) $$\phi \wedge \psi$$ - assumption

3) $$\phi \wedge \psi \rightarrow \phi$$ - axiom

4) $$\phi$$ - Modus Ponens(2,3)

5) $$\psi \rightarrow \theta$$ - Modus Ponens(4,1)

6) $$\phi \wedge \psi \rightarrow\psi$$ - axiom

7) $$\psi$$ - Modus Ponens(6,2)

8) $$\theta$$ - Modus Ponens (5,7)

And after that I applied $$\textit{Deduction theorem}$$.

• Good! ... except watch the numbers .. 5 follows from 4 and 1 ... 7 from 6 and 2 ... and 8 from 5 and 7 – Bram28 Nov 11 '18 at 14:42
• It looks like mostly the right idea, but shouldn’t the lines like “$\phi\land \psi\vdash \phi$ (axiom)” be “$\phi\land\psi \to \phi$ (axiom)” instead? – spaceisdarkgreen Nov 11 '18 at 14:44
• @spaceisdarkgreen en.wikipedia.org/wiki/… – Aleksandra Nov 11 '18 at 14:45
• ... so @spaceisdarkgreen seems to be right ... the axioms in the link you just gave are expressed using $\rightarrow$'s, rather than $\vdash$'s. BTW: whenever you post a question about a formal proof, you should always let us know what rules you are using.. there are many variant proof systems with slightly different rules. – Bram28 Nov 11 '18 at 14:46