Prove that is a tautology without using truth table

Question

I can't find a proper formula to prove that
$$(p\rightarrow q) \rightarrow ((r ∨ p) \rightarrow (r ∨ q)) $$ Is a tautology; considering it's almost exclusively made up of implications. Can somebody help me?

just I want to know how to simplify it.

I tried this.

Answer 1

One approach is to rewrite in conjunctive normal form by replacing $A \implies B$ with $\lnot A \lor B$, pushing negation inward, and distributing $\lor$ over $\land$:

$(p \implies q) \implies ((r \lor p) \implies (r \lor q))$

$\lnot (\lnot p \lor q) \lor (\lnot (r \lor p) \lor (r \lor q))$

$(p \land \lnot q) \lor ((\lnot r \land \lnot p) \lor (r \lor q))$

$(p \land \lnot q) \lor (\lnot r \land \lnot p) \lor r \lor q$

$(p \lor \lnot r \lor r \lor q) \land (p \lor \lnot p \lor r \lor q) \land (\lnot q \lor \lnot r \lor r \lor q) \land (\lnot q \lor \lnot p \lor r \lor q)$

Now notice that each clause contains a pair $A \lor \lnot A$, yielding

$T \land T \land T \land T$, which reduces to $T$.

Answer 2

The given statement is false iff $p\to q=p'\lor q=1\tag1$ $r\lor p=1\tag2$ $r\lor q=0\tag 3$ From $(3),$ $r=q=0.$ From $(2),$ $p=1.$ But then $p'\lor q=0$ contradicting $(1).$

Answer 3

1. Assuming: $p\rightarrow q$
2. Disjunction introduction: $q \rightarrow (r \lor q)$
3. Transitive property of implication (from 1, 2): $p \rightarrow (r \lor q)$
4. Disjunction introduction: $r \rightarrow (r \lor q)$
5. Disjunction elimination (from 3,4): $(r \lor p) \rightarrow (r \lor q)$

Hence assuming 1 gives 5, i.e. $(p\rightarrow q) \rightarrow ((r \lor p) \rightarrow (r \lor q))$, as required

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Let $A$ be $p$, $B$ be $r$ and $C$ be $(r \lor q)$. Statement 5 relies on statements 3 and 4 which respectively are $A \rightarrow C$ and $B \rightarrow C$, and thus indicates that $B$ or $A$ proves $C$, i.e. $(B \lor A) \rightarrow C$