document.write( "Question 1150010: Is (p v q) → (~q → p) a tautology? \n" ); document.write( "
Algebra.Com's Answer #771393 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "p and q are any truth value statements. In other words, they are a variable that holds T or F
\n" ); document.write( "T = true
\n" ); document.write( "F = false\r
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\n" ); document.write( "\n" ); document.write( "Start with a table showing off the various truth value combinations of p and q
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pq
TT
TF
FT
FF
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\n" ); document.write( "\n" ); document.write( "Then add on a ~q column which is the complete opposite of what the q column shows (true flips to false, and vice versa)
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pq~q
TTF
TFT
FTF
FFT

\n" ); document.write( "We'll use this column later, but for now we'll add on a p v q column next.\r
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\n" ); document.write( "\n" ); document.write( "The logical disjunction of two truth values is only false when both pieces are false. In other words, p v q = F when both p = F and q = F at the same time. Otherwise, p v q = T
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pq~qp v q
TTFT
TFTT
FTFT
FFTF
\r
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\n" ); document.write( "\n" ); document.write( "Next we'll have a column for ~q --> p. This conditional is only false when ~q = T and p = F. So if we had T --> F, then that whole thing is false. Otherwise, the statement is true.
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pq~qp v q~q -> p
TTFTT
TFTTT
FTFTT
FFTFF
\r
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\n" ); document.write( "\n" ); document.write( "Finally, the last step is to combine the columns p v q and ~q --> p
\n" ); document.write( "Let A = p v q and B = ~q --> p. The format we want is A --> B
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pq~qp v q~q -> p(p v q) -> (~q -> p)
TTFTTT
TFTTTT
FTFTTT
FFTFFT

\n" ); document.write( "which is what the full completed truth table looks like\r
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\n" ); document.write( "\n" ); document.write( "Note the last column has nothing but T. Each possible outcome leads to (p v q) --> (~q --> p) being a true statement. \r
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\n" ); document.write( "\n" ); document.write( "Answer: Yes it is a tautology\r
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