Question 361802
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Start with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \neg(A\ \large{\vee}\LARGE\ B)\ \Rightarrow\ \neg A \large{\wedge}\LARGE\ \neg B]


From which we can deduce both:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \neg A]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \neg B]


But since:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \neg C\ \Leftrightarrow\ A]


we know that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \neg A\ \Leftrightarrow\ C]


And since:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (F\ \large{\wedge}\LARGE\ H)\ \large{\vee}\LARGE\ \neg C]


we are certain


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (F\ \large{\wedge}\LARGE\ H)]


From which we can deduce


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ F]


But we already demonstrated


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \neg B]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ F\ \large{\wedge}\LARGE\ \neg B]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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