Is this $p \to (q \to p)$ a tautology?

Question

Is $P \to (Q \to P)$ a tautology?

I undertand this:

$P \to (Q \to P)$

$(V \to F = F)$

$(F \to V = V)$

$(F \to F = V)$

$(V \to V = V)$

Why this expression is a tautology, if, $V \to F$ is not true?

turth table:

P | $P \to (Q \to P)$ | R
V | $(V \to F = F)$   | F
V | $(F \to V = V)$   | V
V | $(F \to F = V)$   | V
V | $(V \to V = V)$   | V

Answer 1

Here is the relevant truth table: $$\boxed{\begin{array}{l:l|l|l} q & p & q\to p & p\to(q\to p)\\\hline \sf V & \sf V & \sf V & \sf V \\ \sf F & \sf V & \sf V & \sf V \\ \sf V & \sf F & \sf F & \sf V \\ \sf F & \sf F & \sf V & \sf V \end{array}}$$ so, although $\sf V\to F$ is false, it is true that $\sf F\to(V\to F)\\F\to F\\V$.

Answer 2

Let $\Gamma$ be a set of hypotheses. Let us write, for a sentence $P$, $\Gamma \vdash P$ for "I believe that one can prove $P$ under the assumptions $\Gamma$". You should agree that for any $P$, $Q$, $\Gamma$, if $\Gamma \cup \{P\} \vdash Q$, then $\Gamma \vdash (P \Rightarrow Q)$ (this is what $\Rightarrow$ means), and that for any $P$, $\Gamma \cup \{P\} \vdash P$ (you can assert an hypothesis).

Let us prove that $\emptyset \vdash P \Rightarrow (Q \Rightarrow P)$ :

1) $\{P,Q\} \vdash P$ from the second rule.

2) $\{P\} \vdash Q\Rightarrow P$ from 1) and the first rule.

3) $\emptyset \vdash P \Rightarrow (Q \Rightarrow P)$ from 2) and the first rule.

Answer 3

Note that $q\to p$ is equivalent to $\neg q \lor p$. So your expression is equivalent to

$$\neg p \lor (\neg q \lor p).$$

One of $p$ and $\neg p$ has to be true.

Answer 4

Yes, because $q\to p$ is equivalent to $p\lor\neg q$.