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Metamath Proof Explorer

Theorem bj-spst

Description: Closed form of sps . Once in main part, prove sps and spsd from it. (Contributed by BJ, 20-Oct-2019)

		Ref	Expression
	Assertion	bj-spst	$⊢ (φ \to ψ) \to (\forall x φ \to ψ)$

Step	Hyp	Ref	Expression
1		sp	$⊢ \forall x φ \to φ$
2	1	imim1i	$⊢ (φ \to ψ) \to (\forall x φ \to ψ)$