Is this valid or invalid? prove it with a truth table.

Question

p -> q
q -> p

∴ p ^ q

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Answer 1 · 2018-07-13T22:32:41

Invalid.

Remember the truth table of the mathematical implication, #p->q#, for two propositions #p# and #q#

In other words, it is only false if a true statement implies a false one.

The truth table should look like this:

We see that, if

#{(nu[p]=0),(nu[q]=0) :} => {(nu[p->q]=1),(nu[q->p]=1) :}#

But,

#{(nu[p]=0),(nu[q]=0) :} => nu[p^^q]=0#

where #nu["statement"]# is the truth value of that mathematical statement.

As such, the statement is invalid.