Peter - Math tutor - Auckland
Peter - Math tutor - Auckland

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Peter

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Hat (๐‘ โˆง ๐‘ž)โ‰กT, must be that ๐‘โ‰กT and ๐‘žโ‰กT, hence [(๐‘ โˆง ๐‘ž) โ†’ ๐‘Ÿ]โ‰ก ๐‘Ÿ, and [[(๐‘ โˆง ๐‘ž) โ†’ ๐‘Ÿ] โ†’ ๐‘ ]โ‰ก ๐‘Ÿ โ†’ Fโ‰กยฌ๐‘Ÿ. Secondly, [(๐‘ โ†’ ๐‘Ÿ) โ†’ ๐‘ ] from t

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Therefore proving that [[(๐‘ โˆง ๐‘ž) โ†’ ๐‘Ÿ] โ†’ ๐‘ ] โ†’ [(๐‘ โ†’ ๐‘Ÿ) โ†’ ๐‘ ] โ‰ก T. Q.E.DTherefore proving that [[(๐‘ โˆง ๐‘ž) โ†’ ๐‘Ÿ] โ†’ ๐‘ ] โ†’ [(๐‘ โ†’ ๐‘Ÿ) โ†’ ๐‘ ] โ‰ก T. Q.E.DTherefore proving that [[(๐‘ โˆง ๐‘ž) โ†’ ๐‘Ÿ] โ†’ ๐‘ ] โ†’ [(๐‘ โ†’ ๐‘Ÿ) โ†’ ๐‘ ] โ‰ก T. Q.E.D

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herefore ๐‘ โˆจ (๐‘ž โˆจ ๐‘Ÿ)โ‰กF. So for the left side, since no matter if it is ๐‘โ‰กF or
(๐‘ž โˆจ ๐‘Ÿ)โ‰กF, one out of the two possibility being false is certain, so necessarily ๐‘ โˆง (๐‘ž โˆจ ๐‘Ÿ)โ‰กF, and since F Fโ‰กT, [๐‘ โˆง (๐‘ž โˆจ ๐‘Ÿ)] [(๐‘ โˆง ๐‘ž) โˆจ (๐‘ โˆง ๐‘Ÿ)] is true.

And consider the second case where (๐‘ โˆง ๐‘ž) โˆจ (๐‘ โˆง ๐‘Ÿ)โ‰กT, for it to be held true, necessarily (๐‘ โˆง ๐‘ž)โ‰กT or (๐‘ โˆง ๐‘Ÿ)โ‰กT. Hence either ๐‘ โˆง (๐‘ž โˆง ๐‘Ÿ) โ‰กT or ๐‘ โˆง (๐‘ž โˆง ๐‘Ÿ) โ‰กF. Thus for ๐‘ โˆง (๐‘ž โˆจ ๐‘Ÿ), its truth value must be T.

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