hold on! isnt it making things more complicated?! if we know ¬q =T that means that q=F and then we can see p->(q v r) such as: (p->(F v r)) =T (its given that its true) and then p->r = T in order for p->(F v r) to be true as needed, either p=F and r=F meaning F->F=T, or p=T and r=T meaning T->T=T, either way it doesnt.
help please I was asked in my assignment to proof this in two methods.. p→¬q (q∧r)→¬s r∧s ∴¬p and i did use truth table and it is invalid how am i supposed to proof that in rules of inference ?
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hold on! isnt it making things more complicated?!
if we know ¬q =T that means that q=F and then we can see p->(q v r) such as: (p->(F v r)) =T (its given that its true) and then p->r = T
in order for p->(F v r) to be true as needed, either p=F and r=F meaning F->F=T, or p=T and r=T meaning T->T=T, either way it doesnt.
Thank you!
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help please
I was asked in my assignment to proof this in two methods..
p→¬q
(q∧r)→¬s
r∧s
∴¬p
and i did use truth table and it is invalid
how am i supposed to proof that in rules of inference ?
please I need the book
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