Proof by contradiction.

**albee** · September 26th 2007, 09:07 AM

Use Boolean algebra to show that ((p^q) implies (p implies q)) is a contradiction. do not use truth tables, thanks.
different question!!

**dgomes** · August 20th 2011, 04:14 PM

Are you sure this expression is a contradiction? For me it is a tautology!

**terrorsquid** · August 20th 2011, 10:52 PM

I get a tautology too:

$((p\wedge q) \longrightarrow (p \longrightarrow q))$

$\equiv (-(p\wedge q)\vee (-p \vee q))$

$\equiv (-p\vee-q)\vee (-p\vee q)$

$\equiv -p\vee T$

$\equiv T$

Therefore it's a tautology.

**dgomes** · August 21st 2011, 11:54 AM

That is exactly what I found!

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