Theorem cbvsbdavw

Description: Change bound variable in proper substitution. Deduction form. (Contributed by GG, 14-Aug-2025)

	Ref	Expression
Hypothesis	cbvsbdavw.1	⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( 𝜓 ↔ 𝜒 ) )
Assertion	cbvsbdavw	⊢ ( 𝜑 → ( [ 𝑧 / 𝑥 ] 𝜓 ↔ [ 𝑧 / 𝑦 ] 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		cbvsbdavw.1	⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( 𝜓 ↔ 𝜒 ) )
2		equequ1	⊢ ( 𝑥 = 𝑦 → ( 𝑥 = 𝑡 ↔ 𝑦 = 𝑡 ) )
3	2	adantl	⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( 𝑥 = 𝑡 ↔ 𝑦 = 𝑡 ) )
4	3 1	imbi12d	⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( ( 𝑥 = 𝑡 → 𝜓 ) ↔ ( 𝑦 = 𝑡 → 𝜒 ) ) )
5	4	cbvaldvaw	⊢ ( 𝜑 → ( ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜓 ) ↔ ∀ 𝑦 ( 𝑦 = 𝑡 → 𝜒 ) ) )
6	5	imbi2d	⊢ ( 𝜑 → ( ( 𝑡 = 𝑧 → ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜓 ) ) ↔ ( 𝑡 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑡 → 𝜒 ) ) ) )
7	6	albidv	⊢ ( 𝜑 → ( ∀ 𝑡 ( 𝑡 = 𝑧 → ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜓 ) ) ↔ ∀ 𝑡 ( 𝑡 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑡 → 𝜒 ) ) ) )
8		df-sb	⊢ ( [ 𝑧 / 𝑥 ] 𝜓 ↔ ∀ 𝑡 ( 𝑡 = 𝑧 → ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜓 ) ) )
9		df-sb	⊢ ( [ 𝑧 / 𝑦 ] 𝜒 ↔ ∀ 𝑡 ( 𝑡 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑡 → 𝜒 ) ) )
10	7 8 9	3bitr4g	⊢ ( 𝜑 → ( [ 𝑧 / 𝑥 ] 𝜓 ↔ [ 𝑧 / 𝑦 ] 𝜒 ) )