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Metamath Proof Explorer

Theorem 2uasban

Description: Distribute the unabbreviated form of proper substitution in and out of a conjunction. (Contributed by Alan Sare, 31-May-2014) (Proof modification is discouraged.) (New usage is discouraged.)

		Ref	Expression
	Assertion	2uasban	$⊢ \exists x \exists y ((x = u \land y = v) \land (φ \land ψ)) \leftrightarrow (\exists x \exists y ((x = u \land y = v) \land φ) \land \exists x \exists y ((x = u \land y = v) \land ψ))$

Proof

Step	Hyp	Ref	Expression
1		biid	$⊢ (\exists x \exists y ((x = u \land y = v) \land φ) \land \exists x \exists y ((x = u \land y = v) \land ψ)) \leftrightarrow (\exists x \exists y ((x = u \land y = v) \land φ) \land \exists x \exists y ((x = u \land y = v) \land ψ))$
2	1	2uasbanh	$⊢ \exists x \exists y ((x = u \land y = v) \land (φ \land ψ)) \leftrightarrow (\exists x \exists y ((x = u \land y = v) \land φ) \land \exists x \exists y ((x = u \land y = v) \land ψ))$