I'm doing an exercise from the book of the following:
- $p \leftrightarrow q$
- $\sim p$
- $(q\land\sim\!r)\lor t$
- $(s\lor t)\to r$
conclusion: $r\land\sim\!q$
I applied the biconditional law on the first premise, but seem to be stuck there. Can you guys push me in the right direction?
From 1st premise: $p ↔ q$, by Bi-conditional elimination we get $q → p$ and using Contraposition we have $¬p → ¬q$.
Thus, by Modus Ponens with 2nd premise: $¬q$.
Using Addition we get: $\lnot q \lor r$, and by De Morgan: $\lnot (q \land \lnot r)$.
Thus, $t$ follows from 3rd premise using Disjunctive Syllogism.
Now the conclusion is straightforward.