Logical equivalent of $p\to(q\to p)$

Question

Is Logical equivalent of $p\to(q\to p)$, $p\to(p\wedge q)$ or $p\to(p\vee q)$? I have a truth table: $$\begin{array}{c|c|c|c} p&q&p\wedge q&p\vee q&q\to p&p\to(q\to p)&p\to(q\wedge p)&p\to(q\vee p)\\\hline T&T&T&T&T&T&T&T\\ T&F&F&T&F\color{red}{(T)}&F\color{red}{(T)}&F&T\\ F&T&F&T&F&F\color{red}{(T)}&F\color{red}{(T)}&F\color{red}{(T)}\\ F&F&F&F&F\color{red}{(T)}&F\color{red}{(T)}&F\color{red}{(T)}&F\color{red}{(T)} \end{array}$$ In red color is actual answer.

Answer 1

I think your misunderstanding stems from the basic implication connective.

\begin{array}{|c|c | c|}\hline p & q &p \to q \\ \hline T &T& T\\ T&F& F\\ F&T&T\\ F&F&T\\ \hline \end{array}

Thus, the statement If $P$, then $Q$ is False only when the premise, $P$, is True, while the conclusion, $Q$, is False.

It's not quite right to think that, because the logical statement $F \to \_\_$ is True, then "false implies truth". We simply don't have the logical inference tools to make statements like that.

We have modus tollens that allows us to conclude that, if we know $P \to Q$ and we also know that $P$ is True, we can conclude that $Q$ is also True.

We also have modus ponens that tells us, given $P \to Q$ while $Q$ is False, we can conclude $P$ must also be False (Otherwise the premise would be true and the conclusion false, contradicting $P \to Q$).

So just remember that symbolic logic only goes so far, with regards to interpreting statements like "$0 = 1 \to$ the Riemann Hypothesis." As a logical statement, this is true, but it has no bearing on the truth value of the Riemann Hypothesis.

Answer 2

You should correct your truth table by adding a $q \to p$ column.

From top down it will read T, T, F, T so that $p \to (q \to p)$ is seen to be a tautology.

Answer 3

It seemed that I learned the wrong implication table. The correct order for $p\to q$ is: $$\begin{array}{c|c|c} p&q&p\to q\\\hline T&F&F\\ F&T&T\\ F&F&T\\ T&T&T\end{array}$$

Answer 4

An implication $A \rightarrow B$ is only false, iff $A$ is true and $B$ is false.

So, the implication $ p \rightarrow (q \rightarrow p)$ can only be false, iff $p$ is true and $(q \rightarrow p)$ is false.

The implication $(q \rightarrow p)$ can only be false, iff $q$ is true and $p$ is false. But $p$ is true.

If $p$ was false, then the entire statement $ p \rightarrow (q \rightarrow p)$ would be true again.

So, this statement is always true (such statements are also called tautologies).

In order to find its equivalent, we know that $p$ must be true and we must look what expression can never be false, if $p$ is true. This is $ q \lor p$, because it is true, iff either $p$ or $q$ is true. The statement $q \land p$ can become false, iff $q$ is false.

Hence, the equavilent is $p \rightarrow (p \lor q)$.

Answer 5

Using a truth table in this case is really unnecessary. You are trying to determine what $$ p\to(q\to p)\tag{1} $$ is equivalent to. Well, consider this: \begin{align} p\to(q\to p) &\equiv \neg p\lor(\neg q\lor p)\tag{$r\to s\equiv \neg r\lor s$}\\[0.5em] &\equiv \underbrace{(\neg p\lor p)}_{\text{always true}}\lor\neg q\tag{associativity of $\lor$}\\[0.5em] &\equiv T\lor \neg q\\[0.5em] &\equiv T. \end{align} Thus, we can see $(1)$ is a tautology. This means $(1)$ will be logically equivalent to whatever other tautology we come across. Now consider your other two statements, $p\to(p\land q)$ and $p\to(p\lor q)$. Is one of these a tautology? The first one, $p\to(p\land q)$, will not be true when $p$ is true and $q$ is false (so this statement is not a tautology). However, consider the following for the second statement: \begin{align} p\to(p\lor q) &\equiv \neg p\lor(p\lor q)\\[0.5em] &\equiv (\neg p\lor p)\lor q\\[0.5em] &\equiv T\lor q\\[0.5em] &\equiv T. \end{align} Thus, we can easily see, without use of truth tables, that $$ p\to(q\to p) \equiv p\to(p\lor q). $$

Added: You should note the following: \begin{array}{|c|c|c|}\hline p & q & \neg p & \neg p\lor q & p \to q \\ \hline T & T & F & T & T\\ T & F & F & F & F\\ F & T & T & T & T\\ F & F & T & T & T\\\hline \end{array} That is, memorize the equivalence $$ \Large\color{red}{p\to q\equiv\neg p\lor q} $$ I cannot tell you how many logic/propositional calculus questions I have answered by simple applications of this equivalence. Memorize it. Learn it. Love it. Apply it.