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Metamath Proof Explorer

Theorem sbimd

Description: Deduction substituting both sides of an implication. (Contributed by Wolf Lammen, 24-Nov-2022) Revise df-sb . (Revised by Steven Nguyen, 9-Jul-2023)

	Ref	Expression
Hypotheses	sbimd.1	$⊢ Ⅎ x φ$
	sbimd.2	$⊢ φ \to (ψ \to χ)$
Assertion	sbimd	$⊢ φ \to ([y/ x] ψ \to [y/ x] χ)$

Proof

Step	Hyp	Ref	Expression
1		sbimd.1	$⊢ Ⅎ x φ$
2		sbimd.2	$⊢ φ \to (ψ \to χ)$
3	1 2	alrimi	$⊢ φ \to \forall x (ψ \to χ)$
4		spsbim	$⊢ \forall x (ψ \to χ) \to ([y/ x] ψ \to [y/ x] χ)$
5	3 4	syl	$⊢ φ \to ([y/ x] ψ \to [y/ x] χ)$