Re: [其他] 邏輯等價
※ 引述《skyhigh8988 (Aesthetic)》之銘言：
: 抱歉打不出 "or" - -用 ˇ代替
: ____________________________________________________________________________
: 題目:
: (pˇqˇr)^(pˇtˇ-q)^(pˇ-tˇr)
: =pˇ[r^(tˇ-q)]
: 推敲了一小時得不太到想要的結果
: 希望版上有朋友能夠幫忙一下
: 感謝
Suppose p, then there's nothing to proof. (Both sides are true)
So we may as well assume -p.
Now we need to show (qˇr)^(tˇ-q)^(-tˇr) = r^(tˇ-q).
Again, suppose r, then this is just tˇ-q = tˇ-q, which is true.
So we may as well assume -r.
Now RHS = false, while LHS = q^(tˇ-q)^-t, which is also false,
so the result follows.
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