SOLUTION: Is (p v q) → (~q → p) a tautology?
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Question 1150010: Is (p v q) → (~q → p) a tautology?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
p and q are any truth value statements. In other words, they are a variable that holds T or F
T = true
F = false
Start with a table showing off the various truth value combinations of p and q
Then add on a ~q column which is the complete opposite of what the q column shows (true flips to false, and vice versa)
We'll use this column later, but for now we'll add on a p v q column next.
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
| p | q | ~q | p v q |
| T | T | F | T |
| T | F | T | T |
| F | T | F | T |
| F | F | T | F |
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.
| p | q | ~q | p v q | ~q -> p |
| T | T | F | T | T |
| T | F | T | T | T |
| F | T | F | T | T |
| F | F | T | F | F |
Finally, the last step is to combine the columns p v q and ~q --> p
Let A = p v q and B = ~q --> p. The format we want is A --> B
| p | q | ~q | p v q | ~q -> p | (p v q) -> (~q -> p) |
| T | T | F | T | T | T |
| T | F | T | T | T | T |
| F | T | F | T | T | T |
| F | F | T | F | F | T |
which is what the full completed truth table looks like
Note the last column has nothing but T. Each possible outcome leads to (p v q) --> (~q --> p) being a true statement.
Answer: Yes it is a tautology
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