How to prove this

**MathsNewbie0811** · August 24th 2011, 04:52 AM

How to prove ((p→q)∧p) → q is a tautology using equivalence laws?

**Ackbeet** · August 24th 2011, 05:26 AM

As per this thread, could you please list what equivalences you're allowed to use? What are your inference rules?

**Soroban** · August 24th 2011, 07:57 AM

Hello, MathsNewbie0811!

Prove . $[(p\to q)\wedge p]\:\to\; q$ is a tautology using equivalence laws.

$\begin{array}{ccccccccc} 1. & [( p\to q) \wedge p] \:\to\:q && 1. & \text{Given} \\2. & [(\sim\!p \vee q) \wedge p] \:\to\:q && 2. & \text{Def. Impl'n} \\ 3. & [(\sim\!p \wedge p) \vee (q \wedge p)] \:\to\;q && 3.& \text{Distr.} \\ 4. & [F \vee (q\vee p)] \:\to\:q && 4. & s \:\wedge \sim\!s \:=\:F \\ 5. & \sim(q \wedge p) \vee q && 5. &\text{Def. Impl'n} \\ 6. & (\sim\!q \:\vee \sim\!p) \vee q && 6. & \text{DeMorgan} \\ 7. & (\sim\!q \vee q) \vee p && 7. & \text{Comm, Assoc.} \\ 8. & T \:\vee \sim\!p && 8. & s \:\vee \sim\!s \:=\:T \\ 9. & T && 9. & T \vee s \:=\:T \end{array}$

**terrorsquid** · August 24th 2011, 05:51 PM

Should line 8 be $T \vee p$

Just making sure I understand. From line 7 you apply the fact that $q\vee \lnot q \equiv T$

Why did the p change?

**MathsNewbie0811** · August 24th 2011, 08:12 PM

Thank you.

**emakarov** · August 25th 2011, 01:27 AM

Originally Posted by terrorsquid

Should line 8 be $T \vee p$

Line 8 is correct, but line 7 should be (~q \/ q) \/ ~p.

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