IF P and Q are true and R and S are false,
THEN: < T T F F >
1. ( P & Q & R) -> S is true ( because "(False) -> False" is valid )
and 2. (~P & Q & ~R) -> S is true ( also because "(False) -> False" )
BUT: Q -> S is NOT true ( because "True -> False" is invalid )
The reason that you cannot prove it is because it is not true.
Consider:
IF P and Q are true and R and S are false,
THEN: < T T F F >
1. ( P & Q & R) -> S is true ( because "(False) -> False" is valid )
and 2. (~P & Q & ~R) -> S is true ( also because "(False) -> False" )
BUT: Q -> S is NOT true ( because "True -> False" is invalid )
Therefore it cannot be possible to (validly) derive Q->S from your statements 1 and 2, even if you could use all of the derived rules, replacement, etc.
Pretty hard to prove something that's not true. (In logic anyway :)
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您无法证明它的原因是因为它不真实。
考虑一下：
因此，即使您可以使用所有导出的规则、替换等，也不可能（有效地）从您的陈述 1 和 2 导出 Q->S。
很难证明某些内容不正确。 （无论如何，逻辑上：）
The reason that you cannot prove it is because it is not true.
Consider:
Therefore it cannot be possible to (validly) derive Q->S from your statements 1 and 2, even if you could use all of the derived rules, replacement, etc.
Pretty hard to prove something that's not true. (In logic anyway :)