Question

离散数学试证明 p→q => p→(p∧q) 求高手解答

Answer 1

设p'表示非p,
p→q=p'+q,
同理p→pq=p'+pq,
p→q => p→pq
=(p'+q)'+p'+pq
=pq'+p'+pq
=p(q'+q)+p'
=p+p'
=1,
故命题成立。

Answer 2

p→q
=┐p∨q
=T ∧( ┐p∨q)
=(┐p∨p)∧(┐p∨q)
=┐p∨（p∧q)
=p→（p∧q)