证明:(P->(Q->R))∧(﹁S∨P)∧Q=>(S->R)(1)S P(附加前提)(2)﹁S∨P P(前提)(3)P T(1)(2)I(4)P->(Q->R) P(5)Q->R T(3)(4)I(6)Q P(7)R T(5)(6)I(8)S->R CP规则请解释一下(3)(5)(7)是如何得到的,原式求证明明为
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 12:54:34
![证明:(P->(Q->R))∧(﹁S∨P)∧Q=>(S->R)(1)S P(附加前提)(2)﹁S∨P P(前提)(3)P T(1)(2)I(4)P->(Q->R) P(5)Q->R T(3)(4)I(6)Q P(7)R T(5)(6)I(8)S->R CP规则请解释一下(3)(5)(7)是如何得到的,原式求证明明为](/uploads/image/z/5374519-7-9.jpg?t=%E8%AF%81%E6%98%8E%EF%BC%9A%EF%BC%88P-%3E%28Q-%3ER%29%EF%BC%89%E2%88%A7%EF%BC%88%EF%B9%81S%E2%88%A8P%EF%BC%89%E2%88%A7Q%3D%3E%28S-%3ER%29%281%29S+P%28%E9%99%84%E5%8A%A0%E5%89%8D%E6%8F%90%29%282%29%EF%B9%81S%E2%88%A8P+P%28%E5%89%8D%E6%8F%90%29%283%29P+T%281%29%282%29I%284%29P-%3E%28Q-%3ER%29+P%285%29Q-%3ER+T%283%29%284%29I%286%29Q+P%287%29R+T%285%29%286%29I%288%29S-%3ER+CP%E8%A7%84%E5%88%99%E8%AF%B7%E8%A7%A3%E9%87%8A%E4%B8%80%E4%B8%8B%EF%BC%883%EF%BC%89%EF%BC%885%EF%BC%89%EF%BC%887%EF%BC%89%E6%98%AF%E5%A6%82%E4%BD%95%E5%BE%97%E5%88%B0%E7%9A%84%2C%E5%8E%9F%E5%BC%8F%E6%B1%82%E8%AF%81%E6%98%8E%E6%98%8E%E4%B8%BA)
xRN@|RPi
חrsH_}**wB BT51<]O~E!3ͮ^1hKv;=\2ҒB~@V:7 (}0rkˬ1W;GnKgi+l5|lQ{ )6tEPZXT.Túzj4QÓ*w/ưh[q]X7~C9r\Y{{qs
Ĥ Cz
5tLC^)gU2윋ƭp.8@aH,_47ՐM=W171 &u?6w/ȳ RZeT1Ьp,\N4g qE/DCU>;7c@
j;?¹$ɑREd*
:3|
=QH^\/6k*vpم)#a
证明:(P->(Q->R))∧(﹁S∨P)∧Q=>(S->R)(1)S P(附加前提)(2)﹁S∨P P(前提)(3)P T(1)(2)I(4)P->(Q->R) P(5)Q->R T(3)(4)I(6)Q P(7)R T(5)(6)I(8)S->R CP规则请解释一下(3)(5)(7)是如何得到的,原式求证明明为
证明:(P->(Q->R))∧(﹁S∨P)∧Q=>(S->R)
(1)S P(附加前提)
(2)﹁S∨P P(前提)
(3)P T(1)(2)I
(4)P->(Q->R) P
(5)Q->R T(3)(4)I
(6)Q P
(7)R T(5)(6)I
(8)S->R CP规则
请解释一下(3)(5)(7)是如何得到的,原式求证明明为∧符号,而(3)(5)(7)是若推出只能是∨符号的条件下
求指教~
证明:(P->(Q->R))∧(﹁S∨P)∧Q=>(S->R)(1)S P(附加前提)(2)﹁S∨P P(前提)(3)P T(1)(2)I(4)P->(Q->R) P(5)Q->R T(3)(4)I(6)Q P(7)R T(5)(6)I(8)S->R CP规则请解释一下(3)(5)(7)是如何得到的,原式求证明明为
∧说明左边的三个条件都是成立的
﹁S∨P成立,又由S成立给定,那么﹁S不成立,只能P成立
所以就是由(1)(2)能得到P成立
因为P->(Q->R)成立,P又成立,所以Q->R成立
以此类推
我看不懂,我才高一