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Metamath Proof Explorer

Theorem pm5.21nd

Description: Eliminate an antecedent implied by each side of a biconditional. Variant of pm5.21ndd . (Contributed by NM, 20-Nov-2005) (Proof shortened by Wolf Lammen, 4-Nov-2013)

	Ref	Expression
Hypotheses	pm5.21nd.1	$⊢ (φ \land ψ) \to θ$
	pm5.21nd.2	$⊢ (φ \land χ) \to θ$
	pm5.21nd.3	$⊢ θ \to (ψ \leftrightarrow χ)$
Assertion	pm5.21nd	$⊢ φ \to (ψ \leftrightarrow χ)$

Proof

Step	Hyp	Ref	Expression
1		pm5.21nd.1	$⊢ (φ \land ψ) \to θ$
2		pm5.21nd.2	$⊢ (φ \land χ) \to θ$
3		pm5.21nd.3	$⊢ θ \to (ψ \leftrightarrow χ)$
4	1	ex	$⊢ φ \to (ψ \to θ)$
5	2	ex	$⊢ φ \to (χ \to θ)$
6	3	a1i	$⊢ φ \to (θ \to (ψ \leftrightarrow χ))$
7	4 5 6	pm5.21ndd	$⊢ φ \to (ψ \leftrightarrow χ)$