Are these 2 equivalent?
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05-11-2019 - |
質問
Is ∀x∀y∀z[φ(x,y)∧p(y,z)->p(x,z)] equivalent to ∀x∀y∀z[φ(x,y)∧p(x,z)->p(y,z)] ?
The only thing I can think of is that this question can be answered if we show that p->q
is equals (↔) το q->p
, but that's not true because p->q ↔ ¬p->¬q
, hence p->q
is not equals to q->p
. However, I do not know if my logic is correct and if it can be accepted as an answer.
I also think that the (first) ∀x∀y∀z[φ(x,y)∧ part is irrelevant and that i have to focus only to the last part.
正しい解決策はありません
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