Show that (p → q) ∧ (q → r) → (p → r) is a tautology
The truth table ((p→Q)∧(Q→R))→(P→R):((p\to Q)\land (Q\to R))\to (P\to R):((p→Q)∧(Q→R))→(P→R):
Therefore ((p→Q)∧(Q→R))→(P→R):((p\to Q)\land (Q\to R))\to (P\to R):((p→Q)∧(Q→R))→(P→R):
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