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Metamath Proof Explorer

Theorem sbbidv

Description: Deduction substituting both sides of a biconditional, with ph and x disjoint. See also sbbid . (Contributed by Wolf Lammen, 6-May-2023) (Proof shortened by Steven Nguyen, 6-Jul-2023)

		Ref	Expression
	Hypothesis	sbbidv.1	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
	Assertion	sbbidv	⊢ ( 𝜑 → ( [ 𝑡 / 𝑥 ] 𝜓 ↔ [ 𝑡 / 𝑥 ] 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		sbbidv.1	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
2	1	alrimiv	⊢ ( 𝜑 → ∀ 𝑥 ( 𝜓 ↔ 𝜒 ) )
3		spsbbi	⊢ ( ∀ 𝑥 ( 𝜓 ↔ 𝜒 ) → ( [ 𝑡 / 𝑥 ] 𝜓 ↔ [ 𝑡 / 𝑥 ] 𝜒 ) )
4	2 3	syl	⊢ ( 𝜑 → ( [ 𝑡 / 𝑥 ] 𝜓 ↔ [ 𝑡 / 𝑥 ] 𝜒 ) )