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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 $\textit{Deduction theorem}$.

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    $\begingroup$ Good! ... except watch the numbers .. 5 follows from 4 and 1 ... 7 from 6 and 2 ... and 8 from 5 and 7 $\endgroup$ – Bram28 Nov 11 '18 at 14:42
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    $\begingroup$ 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? $\endgroup$ – spaceisdarkgreen Nov 11 '18 at 14:44
  • $\begingroup$ @spaceisdarkgreen en.wikipedia.org/wiki/… $\endgroup$ – Aleksandra Nov 11 '18 at 14:45
  • $\begingroup$ ... 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. $\endgroup$ – Bram28 Nov 11 '18 at 14:46

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