Show that $ \{ (\phi \wedge \psi) \rightarrow \theta \} \vdash \phi \rightarrow(\psi \rightarrow \theta)$

Question

How to show that $ \{ (\phi \wedge \psi) \rightarrow \theta \} \vdash \phi \rightarrow(\psi \rightarrow \theta)$?

I tried to do it using deduction theorem and got $\vdash((\phi \wedge \psi) \rightarrow \theta) \rightarrow (\phi \rightarrow(\psi \rightarrow \theta))$. After that rewrote it like $\vdash(\neg(\phi \rightarrow \neg \psi) \rightarrow \theta) \rightarrow (\phi \rightarrow(\psi \rightarrow \theta))$, but now I dont have no idea what to do next...

Answer 1

Hint

We can use the following axiom system for propositional calculus, with modus ponens as the only rule of inference.

As usual, Ax.1 and Ax.2 are used to prove the Deduction Theorem.

1) $(ϕ∧ψ)→θ$ --- premise

2) $ϕ$ --- assumption [a]

3) $ψ$ --- assumption [b]

4) $\vdash ϕ \to (ψ \to (ϕ∧ψ))$ --- axiom

5) $(ϕ∧ψ)$ --- from 2) and 3) and 4) by modus ponens twice

6) $θ$ --- from 5) and 1) by modus ponens

7) $(ψ \to θ)$ --- from 3) and 6) by Deduction Th, discharging temporary assumption [b]

8) $ϕ \to (ψ \to θ)$ --- from 2) and 7) by Deduction Th, discharging temporary assumption [a].