Question

Show that the given conditional statement ¬ (p → q) → p is a tautology

Answer 1

We know that,

$\displaystyle\sf{\longrightarrow p\implies q \equiv \lnot p\lor q}$

Thus,

$\displaystyle\sf{\longrightarrow \lnot(p\implies q)\implies p\equiv\lnot (\lnot p\lor q)\implies p}$

Taking $\displaystyle\sf {\lnot p\lor q=r,}$

$\displaystyle\sf{\longrightarrow \lnot(p\implies q)\implies p\equiv\lnot r\implies p}$

$\displaystyle\sf{\longrightarrow \lnot(p\implies q)\implies p\equiv r\lor p}$

$\displaystyle\sf{\longrightarrow \lnot(p\implies q)\implies p\equiv (\lnot p\lor q)\lor p}$

$\displaystyle\sf{\longrightarrow \lnot(p\implies q)\implies p\equiv \lnot p\lor q\lor p}$

$\displaystyle\sf{\longrightarrow \lnot(p\implies q)\implies p\equiv (\lnot p\lor p)\lor q}$

The statement contains $\displaystyle\sf {\lnot p\lor p}$ which is always true for any truth value of $\displaystyle\sf {p,}$ whether $\displaystyle\sf {p}$ is true or false, and no matter what the truth value of $\displaystyle\sf {q}$ is.

This implies our statement is a tautology.

Hence Proved!

Answer 2

GIVEN:

• ¬ (p → q) → p

TO PROVE:

• ¬ (p → q) → p is a tautology

SOLUTION:

Here,

we know that,

$\tt{↣ p\implies q \equiv \lnot p\lor q}$

Thus,

$\tt{↣ \lnot(p\implies q)\implies p\equiv\lnot (\lnot p\lor q)\implies p}$

$\tt{ ↣now \: take \lnot p\lor q=r,}$

$\tt{↣ \lnot(p\implies q)\implies p\equiv\lnot r\implies p}$

$\tt{↣ \lnot(p\implies q)\implies p\equiv r\lor p}$

$\tt{↣ \lnot(p\implies q)\implies p\equiv (\lnot p\lor q)\lor p}$

$\tt{↣ \lnot(p\implies q)\implies p\equiv \lnot p\lor q\lor p}$

$\tt{↣\lnot(p\implies q)\implies p\equiv (\lnot p\lor p)\lor q}$

$\tt{here \: The \: statement \: contains \lnot p\lor p}$

$\tt{\: which \: is \: true \: forever \: for \: any \: true \: value}$

$\tt{of \: p, or \: whether \: p \: is \: true \: or \: false, no }$

$\tt{matter \: what \: the \: truth \: value \: of \: q \: is.}$

So, this implies that our statement is a tautology.