用“p→q=~p∨q”证明:(p→q)∧(q→r)=> p→r
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/29 02:18:09
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用“p→q=~p∨q”证明:(p→q)∧(q→r)=> p→r
用“p→q=~p∨q”证明:(p→q)∧(q→r)=> p→r
用“p→q=~p∨q”证明:(p→q)∧(q→r)=> p→r
证:(p→q)∧(q→r)
=(~p∨q)∧(~q∨r)
=[~p∧(~q∨r)]∨[q∧(~q∨r)]
=[(~p∧~q)∨(~p∧r)]∨[(q∧~q)∨(q∧r)]
=(~p∧~q)∨(~p∧r)∨0∨(q∧r)
=(~p∧~q)∨(~p∧r)∨(q∧r)
=(~p∧~q)∨[(~p∨q)∧r]
={~p∨[(~p∨q)∧r]}∧{~q∨[(~p∨q)∧r]}
={[~p∨(~p∨q)]∧(~p∨r)}∧{[~q∨(~p∨q)]∧(~q∨r)}
=[(~p∨q)∧(~p∨r)]∧1∧(~q∨r)
=[(~p∨q)∧(~p∨r)]∧(~q∨r)
=[(~p∨q)∧(~q∨r)]∧(~p∨r)
=[(p→q)∧(q→r)]∧( p→r)
即有
(p→q)∧(q→r)=[(p→q)∧(q→r)]∧( p→r)
所以
[(p→q)∧(q→r)]}∨( p→r)
{[(p→q)∧(q→r)]∧( p→r)}∨( p→r)
={~[(p→q)∧(q→r)]}∨[~( p→r)]∨( p→r)
=1
即(p→q)∧(q→r)=> p→r 恒成立 证毕
证:(p→q)∧(q→r)
=(~p∨q)∧(~q∨r)
=[~p∧(~q∨r)]∨[q∧(~q∨r)]
=[(~p∧~q)∨(~p∧r)]∨[(q∧~q)∨(q∧r)]
=(~p∧~q)∨(~p∧r)∨0∨(q∧r)
=(~p∧~q)∨(~p∧r)∨(q∧r)
=(~p∧~q)∨[(~p∨q)∧r]
={~p∨[(~p∨q)∧r]}∧{~q...
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证:(p→q)∧(q→r)
=(~p∨q)∧(~q∨r)
=[~p∧(~q∨r)]∨[q∧(~q∨r)]
=[(~p∧~q)∨(~p∧r)]∨[(q∧~q)∨(q∧r)]
=(~p∧~q)∨(~p∧r)∨0∨(q∧r)
=(~p∧~q)∨(~p∧r)∨(q∧r)
=(~p∧~q)∨[(~p∨q)∧r]
={~p∨[(~p∨q)∧r]}∧{~q∨[(~p∨q)∧r]}
={[~p∨(~p∨q)]∧(~p∨r)}∧{[~q∨(~p∨q)]∧(~q∨r)}
=[(~p∨q)∧(~p∨r)]∧1∧(~q∨r)
=[(~p∨q)∧(~p∨r)]∧(~q∨r)
=[(~p∨q)∧(~q∨r)]∧(~p∨r)
=[(p→q)∧(q→r)]∧( p→r)
即有
(p→q)∧(q→r)=[(p→q)∧(q→r)]∧( p→r)
所以
{~[(p→q)∧(q→r)]}∨( p→r)
=~{[(p→q)∧(q→r)]∧( p→r)}∨( p→r)
={~[(p→q)∧(q→r)]}∨[~( p→r)]∨( p→r)
=1
收起
证:(p→q)∧(q→r)
=(~p∨q)∧(~q∨r)
=[~p∧(~q∨r)]∨[q∧(~q∨r)]
=[(~p∧~q)∨(~p∧r)]∨[(q∧~q)∨(q∧r)]
=(~p∧~q)∨(~p∧r)∨0∨(q∧r)d=(~p∧~q)∨(~p∧r)∨(q∧r)
=(~p∧~q)∨[(~p∨q)∧r]
={~p∨[(~p∨q)∧r]}∧{~q∨[(~...
全部展开
证:(p→q)∧(q→r)
=(~p∨q)∧(~q∨r)
=[~p∧(~q∨r)]∨[q∧(~q∨r)]
=[(~p∧~q)∨(~p∧r)]∨[(q∧~q)∨(q∧r)]
=(~p∧~q)∨(~p∧r)∨0∨(q∧r)d=(~p∧~q)∨(~p∧r)∨(q∧r)
=(~p∧~q)∨[(~p∨q)∧r]
={~p∨[(~p∨q)∧r]}∧{~q∨[(~p∨q)∧r]}
={[~p∨(~p∨q)]∧(~p∨r)}∧{[~q∨(~p∨q)]∧(~q∨r)}
=[(~p∨q)∧(~p∨r)]∧1∧(~q∨r)
=[(~p∨q)∧(~p∨r)]∧(~q∨r)
=[(~p∨q)∧(~q∨r)]∧(~p∨r)w:(p→q)∧(q→r)
=(~p∨q)∧(~q∨r)
=[~p∧(~q∨r)]∨[q∧(~q∨r)]
=[(~p∧~q)∨(~p∧r)]∨[(q∧~q)∨(q∧r)]
=(~p∧~q)∨(~p∧r)∨0∨(q∧r)
=(~p∧~q)∨(~p∧r)∨(q∧r)
=(~p∧~q)∨[(~p∨q)∧r]
={~p∨[(~p∨q)∧r]}∧{~q∨[(~p∨q)∧r]}
={[~p∨(~p∨q)]∧(~p∨r)}∧{[~q∨(~p∨q)]∧(~q∨r)}
=[(~p∨q)∧(~p∨r)]∧1∧(~q∨r)
=[(~p∨q)∧(~p∨r)]∧(~q∨r)
=[(~p∨q)∧(~q∨r)]∧(~p∨r)
=[(p→q)∧(q→r)]∧( p→r)
即有
(p→q)∧(q→r)=[(p→q)∧(q→r)]∧( p→r)
=[(p→q)∧(q→r)]∧( p→r)
即有
(p→q)∧(q→r)=[(p→q)∧(q→r)]∧( p→r)
所以
{~[(p→q)∧(q→r)]}∨( p→r)
=~{[(p→q)∧(q→r)]∧( p→r)}∨( p→r)
={~[(p→q)∧(q→r)]}∨[~( p→r)]∨( p→r)
=1
即(p→q)∧(q→r)=> p→r 恒成立 证毕
收起