How to prove ((p→q)∧p) → q is a tautology using equivalence laws?
As per this thread, could you please list what equivalences you're allowed to use? What are your inference rules?
Hello, MathsNewbie0811!
Prove .$\displaystyle [(p\to q)\wedge p]\:\to\; q$ is a tautology using equivalence laws.
$\displaystyle \begin{array}{ccccccccc} 1. & [( p\to q) \wedge p] \:\to\:q && 1. & \text{Given} \\2. & [(\sim\!p \vee q) \wedge p] \:\to\:q && 2. & \text{Def. Impl'n} \\ 3. & [(\sim\!p \wedge p) \vee (q \wedge p)] \:\to\;q && 3.& \text{Distr.} \\ 4. & [F \vee (q\vee p)] \:\to\:q && 4. & s \:\wedge \sim\!s \:=\:F \\ 5. & \sim(q \wedge p) \vee q && 5. &\text{Def. Impl'n} \\ 6. & (\sim\!q \:\vee \sim\!p) \vee q && 6. & \text{DeMorgan} \\ 7. & (\sim\!q \vee q) \vee p && 7. & \text{Comm, Assoc.} \\ 8. & T \:\vee \sim\!p && 8. & s \:\vee \sim\!s \:=\:T \\ 9. & T && 9. & T \vee s \:=\:T \end{array}$