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Metamath Proof Explorer

Theorem imdistand

Description: Distribution of implication with conjunction (deduction form). (Contributed by NM, 27-Aug-2004)

		Ref	Expression
	Hypothesis	imdistand.1	$⊢ φ \to (ψ \to (χ \to θ))$
	Assertion	imdistand	$⊢ φ \to ((ψ \land χ) \to (ψ \land θ))$

Step	Hyp	Ref	Expression
1		imdistand.1	$⊢ φ \to (ψ \to (χ \to θ))$
2		imdistan	$⊢ (ψ \to (χ \to θ)) \leftrightarrow ((ψ \land χ) \to (ψ \land θ))$
3	1 2	sylib	$⊢ φ \to ((ψ \land χ) \to (ψ \land θ))$