Theorem sbequbidv

Description: Deduction substituting both sides of a biconditional. (Contributed by GG, 1-Sep-2025)

	Ref	Expression
Hypotheses	sbequbidv.1	⊢ ( 𝜑 → 𝑢 = 𝑣 )
	sbequbidv.2	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
Assertion	sbequbidv	⊢ ( 𝜑 → ( [ 𝑢 / 𝑥 ] 𝜓 ↔ [ 𝑣 / 𝑥 ] 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		sbequbidv.1	⊢ ( 𝜑 → 𝑢 = 𝑣 )
2		sbequbidv.2	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
3		equequ2	⊢ ( 𝑢 = 𝑣 → ( 𝑡 = 𝑢 ↔ 𝑡 = 𝑣 ) )
4	1 3	syl	⊢ ( 𝜑 → ( 𝑡 = 𝑢 ↔ 𝑡 = 𝑣 ) )
5	2	imbi2d	⊢ ( 𝜑 → ( ( 𝑥 = 𝑡 → 𝜓 ) ↔ ( 𝑥 = 𝑡 → 𝜒 ) ) )
6	5	albidv	⊢ ( 𝜑 → ( ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜓 ) ↔ ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜒 ) ) )
7	4 6	imbi12d	⊢ ( 𝜑 → ( ( 𝑡 = 𝑢 → ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜓 ) ) ↔ ( 𝑡 = 𝑣 → ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜒 ) ) ) )
8	7	albidv	⊢ ( 𝜑 → ( ∀ 𝑡 ( 𝑡 = 𝑢 → ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜓 ) ) ↔ ∀ 𝑡 ( 𝑡 = 𝑣 → ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜒 ) ) ) )
9		df-sb	⊢ ( [ 𝑢 / 𝑥 ] 𝜓 ↔ ∀ 𝑡 ( 𝑡 = 𝑢 → ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜓 ) ) )
10		df-sb	⊢ ( [ 𝑣 / 𝑥 ] 𝜒 ↔ ∀ 𝑡 ( 𝑡 = 𝑣 → ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜒 ) ) )
11	8 9 10	3bitr4g	⊢ ( 𝜑 → ( [ 𝑢 / 𝑥 ] 𝜓 ↔ [ 𝑣 / 𝑥 ] 𝜒 ) )