Proof $(P \land S) \rightarrow \lnot q$ in hilbert system How can I proof $(P \land S) \rightarrow \lnot q$ using this principles :


$p \rightarrow (q \lor r) $
$q \rightarrow \lnot r $
$r \leftrightarrow s  $


in Hilbert system and modus ponens?
You don't need principle 1, just start with the following propositional tautology:

$\fbox{4}$ $(q \rightarrow \lnot~r)$ $\rightarrow$ $((r \leftrightarrow s)$ $\rightarrow$ $((p \land s) \rightarrow \lnot~q))$ $\color{gray}{(propositional~tautology)}$

$\fbox{5}$ $(r \leftrightarrow s)$ $\rightarrow$ $((p \land s) \rightarrow \lnot~q)$ $\color{gray}{(from~2,4~by~modus~ponens)}$

$\fbox{6}$ $(p \land s) \rightarrow \lnot~q$ $\color{gray}{(from~3,5~by~modus~ponens)}$
Axiom 2 is (q→¬r).  Substituting q with (P $\land$ S) yields 4

4 [(P $\land$ S)→¬r].

Substituting  r with q in 4 yields 5

5 [(P $\land$ S)→¬$\lnot$q].