(p -> r) ^ (r -> q ) and (p -> q)
thanks
Hello, Ammar!
$\displaystyle \text{Are these propositions logically equivalent?}$
. . . $\displaystyle (p \to r) \wedge (r\to q)\,\text{ and }\,(p \to q)$
No, their truth-table values are not identical.
$\displaystyle \begin{array}{|c|c|c || c|c|c|c|c|c|c || c|c|c|} p&q&r & (p & \to & r) & \wedge & (r & \to & q) & p & \to & q \\ \hline T&T&T & T&T&T&{\color{red}T}&T&T&T & T&{\color{red}T}&T \\ T&T&F & T&F&F&{\color{red}F}&F&T&T & T&{\color{red}T}&T \\ T&F&T & T&T&T&{\color{red}F}&T&F&F & T&{\color{red}F}&F \\ T&F&F & T&F&F&{\color{red}F}&F&T&F & T&{\color{red}F}&F \\ F&T&T & F&T&T&{\color{red}T}&T&T&T& F&{\color{red}T}&T \\ F&T&F & F&T&F&{\color{red}T}&F&T&T & F&{\color{red}T}&T \\ F&F&T & F &T&T&{\color{red}F}&T&F&F & F&{\color{red}T}&F \\ F&F&F & F&T&F&{\color{red}T}&F&T&F & F&{\color{red}T}&F \\ \hline &&& 1&2&1 &3& 1&2&1 & 1&2&1 \\ \hline \end{array}$
Do you understand the difference between "A is equivalent to B" and "A implies B"?
"A implies B" says that if A is true then B must be true but says nothing about what happens if A if false.
"A is equivalent to B" says that if A is true then B must be true and if A is false then B is false".
In order that (p->r)^(r->q) be true both p->r and r->q must be true. In that case, if p is true, r is true and then q is true. In other word, if p is true so is q: p->q.
But (p->r)^(r->q) is false one of p->r or r->q must be false. It might be the case, for example that p is true and r is false, so that p->r is false. That tells us nothing about whether or not q is true or false.