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

different question!!:)

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- Sep 26th 2007, 09:07 AMalbeeProof by contradiction.
Use Boolean algebra to show that ((p^q) implies (p implies q)) is a contradiction. do not use truth tables, thanks.

different question!!:) - Aug 20th 2011, 04:14 PMdgomesRe: Proof by contradiction (please help, urgent) :)
Are you sure this expression is a contradiction? For me it is a tautology!

- Aug 20th 2011, 10:52 PMterrorsquidRe: Proof by contradiction.
I get a tautology too:

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

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

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

$\displaystyle \equiv -p\vee T$

$\displaystyle \equiv T$

Therefore it's a tautology. - Aug 21st 2011, 11:54 AMdgomesRe: Proof by contradiction.
That is exactly what I found!