Question

tell me the answer the question is from mathematical logic​

Answer 1

Given,

$\longrightarrow\sf{p\implies (p\lor q)}$

We say "p  implies p or q". We have to find whether the statement is true or false.

We know that "p implies q" is equivalent to "not p or q", i.e.,

$\longrightarrow\sf{p\implies q\ \equiv\ \lnot p\lor q}$

Then we have,

$\longrightarrow\sf{p\implies (p\lor q)\ \equiv\ \lnot p\lor(p\lor q)}$

By associative law,

$\longrightarrow\sf{p\implies (p\lor q)\ \equiv\ (\lnot p\lor p)\lor q}$

But $\sf{\lnot p\lor p}$ is always a true statement. Let,

• $\sf{\lnot p\lor p=T}$

where $\sf{T}$ indicates true statement. Similarly false statement is indicated by $\sf{F.}$

Then,

$\longrightarrow\sf{p\implies (p\lor q)\ \equiv\ T\lor q}$

Whether $\sf{q}$ is true or false, a statement with $\sf{q}$ or a true statement is always true, right?

Hence,

$\longrightarrow\sf{\underline{\underline{p\implies (p\lor q)\ \equiv\ T}}}$

Hence our statement is always true. Let's check by forming a truth table.

$\begin{tabular}{|c|c|c|c|}\cline{1-4}&&&\\\sf{p}&\sf{q}& \mathsf{p\lor q} & \sf{p\implies(p\lor q)} \\&&&\\\cline{1-4}\sf{T}&\sf{T}&\sf{T}&\sf{T}\\\cline{1-4}\sf{T}&\sf{F}&\sf{T}&\sf{T}\\\cline{1-4}\sf{F}&\sf{T}&\sf{T}&\sf{T}\\\cline{1-4}\sf{F}&\sf{F}&\sf{F}&\sf{T}\\\cline{1-4}\end{tabular}$

Hence, $\sf{p\implies (p\lor q)}$ is always a true statement.