$z>1\Rightarrow z>3$
this is a really simple example to help me understand.
So, if $z\in \mathbb{R}$, then when LHS is false, the statement is true.
But if LHS is true AND RHS is true $\Leftrightarrow$ the statement is false.
Is this right?
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2Almost duplicate: In classical logic, why is (p⇒q) True if p is False and q is True? – hmakholm left over Monica Oct 24 '16 at 20:51
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@HenningMakholm yes, i know it is almost surely a duplicate of some question, but i needed some "direct" assistance. If i can myself link to a dupliate i will do (please say the code for here) – SAJW Oct 24 '16 at 20:53
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i voted myself to close it because duplicate. so if nessecary do it too :) (i dont grief about it) – SAJW Oct 24 '16 at 21:40
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The only case when your implication is false, is when the lhs is true, AND the rhs is false – amWhy Oct 25 '16 at 11:09
2 Answers
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No, that's not right: the only way a sentence "$A\implies B$" is false is if $A$ is true and $B$ is false. E.g. the sentence $$1+1=2\implies 3=87$$ is false, but $$3=87\implies 1+1=2$$ is true.
Noah Schweber
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oh man, if i werent wasted i didnt made that mistake, but thanks anyways, thats exactly what i wanted to know :)$\$ i edited my originial question, but your answer is till ok ;) – SAJW Oct 24 '16 at 20:48
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Consider a real life example. Mr. X says," If my brother wears a white shirt then so do I".
Take $p$: Brother of Mr. X wears a white shirt.
$q$: Mr. X wears a white shirt.
Symbolically, it is the conditional $p\implies q$.
On some day, Mr. X must feel guilty of lying about his past statement(i.e. conditional $p\implies q$ is $F$) if he will not be able to wear a white shirt ($q$ is $F$) whenever his brother will be wearing a white shirt ($p$ is $T$).
Nitin Uniyal
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