Lets start by filling in what we have been given.
#[
(P,Q,notP,notQ,(P->notQ), ((P->notQ)vvnotP)),
(T,T,,,,),
(T,F,,,,),
(F,T,,,,),
(F,F,,,,)
]#
Right now we have our two variables, #P# and #Q#, and every possible combination of values. We can fill in the next two columns, #notP# and #notQ#, by switching the values of #P# and #Q# respectively.
#[
(P,Q,notP,notQ,(P->notQ), ((P->notQ)vvnotP)),
(T,T,F,F,,),
(T,F,F,T,,),
(F,T,T,F,,),
(F,F,T,T,,)
]#
Next we can work on the implication column. The statement reads "If #P# then #notQ#" There is only one case where this statement is false. If #P# is true, and #notQ# is false, then you cannot infer #notQ# from #P#. Therefore, the first case is false, and all others are true.
#[
(P,Q,notP,notQ,(P->notQ), ((P->notQ)vvnotP)),
(T,T,F,F,F,),
(T,F,F,T,T,),
(F,T,T,F,T,),
(F,F,T,T,T,)
]#
Now we can solve the last column, which is an or statement. An or statement is true if either condition is true. In this case, we will be referring to the third and fifth column. The only case where neither condition is true is the first case, so that will be false, and the rest will be true.
#[
(P,Q,notP,notQ,(P->notQ), ((P->notQ)vvnotP)),
(T,T,F,F,F,F),
(T,F,F,T,T,T),
(F,T,T,F,T,T),
(F,F,T,T,T,T)
]#
