作者turboho (西卡拉)
看板Math
标题Re: [其他] 逻辑等价
时间Tue Mar 1 12:31:05 2011
※ 引述《turboho (西卡拉)》之铭言：
: ※ 引述《skyhigh8988 (Aesthetic)》之铭言：
: : 抱歉打不出 "or" - -用 ˇ代替
: : ____________________________________________________________________________
: : 题目:
: : (pˇqˇr)^(pˇtˇ-q)^(pˇ-tˇr)
: : =pˇ[r^(tˇ-q)]
: : 推敲了一小时得不太到想要的结果
: : 希望版上有朋友能够帮忙一下
: : 感谢
: 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.
Let's see if this works for you:
(pˇqˇr)^(pˇtˇ-q)^(pˇ-tˇr)
= {pˇ[(qˇr)^(tˇ-q)]}^(pˇ-tˇr)
= pˇ[(qˇr)^(tˇ-q)^(-tˇr)]
= pˇ[(qˇr)^(-tˇr)^(tˇ-q)]
= pˇ{[rˇ(q^-t)]^(tˇ-q)}
= pˇ{[rˇ-(-qˇt)]^(tˇ-q)}
= pˇ[r^(tˇ-q)]
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1F：推 skyhigh8988 :谢谢您 我懂了~ 03/01 12:34