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Metamath Proof Explorer

Theorem imdistanda

Description: Distribution of implication with conjunction (deduction version with conjoined antecedent). (Contributed by Jeff Madsen, 19-Jun-2011)

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

Proof

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