※ 引述《ntddt (哀便毛)》之銘言：
: 請教由
: (8): All A and C and not D are contradictory
: (9): All A and not B and not D are contradictory
: 怎推導出
: (10): All A and not D are B and not C
: 步驟愈詳細愈好, Thanks
(A & C & not D) -> _
(A & not B & not D) -> _
推導 A & not D -> B & not C:
A & not D = (A & not D) & (C | not C) = (A & C & not D) | (A & not C & not D).
因為 (A & C & not D) 是 contradictory,即 A & C & no D |- _,
所以 A & not D = A & not C & not D = (A & not C & not D) & (B | not B)
= (A & B & not C & not D) | (A & not B & not C & not D).
再因為 (A & not B & not D) 是 contradictory,即 A & not B & not C & notD
= _ & not C = _,
所以 A & not D = A & B & not C & not D.
於是 A & not D -> B & not C.
以上不知道有沒有搞錯東西,請指教.
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※ 編輯: yauhh 來自: 59.112.224.131 (04/09 16:59)