Theorem cbvsbdavw 36825

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

Hypothesis

Ref	Expression
cbvsbdavw.1	⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
cbvsbdavw	⊢ (𝜑 → ([𝑧 / 𝑥]𝜓 ↔ [𝑧 / 𝑦]𝜒))

Distinct variable groups:   𝜑,𝑥,𝑦   𝜓,𝑦   𝜒,𝑥

Allowed substitution hints:   𝜑(𝑧)   𝜓(𝑥, 𝑧)   𝜒(𝑦, 𝑧)

Proof of Theorem cbvsbdavw

Dummy variable 𝑡 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		equequ1 2058	. . . . . . 7 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝑡 ↔ 𝑦 = 𝑡))
2	1	adantl 487	. . . . . 6 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝑥 = 𝑡 ↔ 𝑦 = 𝑡))
3		cbvsbdavw.1	. . . . . 6 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝜓 ↔ 𝜒))
4	2, 3	imbi12d 347	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → ((𝑥 = 𝑡 → 𝜓) ↔ (𝑦 = 𝑡 → 𝜒)))
5	4	cbvaldvaw 2071	. . . 4 ⊢ (𝜑 → (∀𝑥(𝑥 = 𝑡 → 𝜓) ↔ ∀𝑦(𝑦 = 𝑡 → 𝜒)))
6	5	imbi2d 343	. . 3 ⊢ (𝜑 → ((𝑡 = 𝑧 → ∀𝑥(𝑥 = 𝑡 → 𝜓)) ↔ (𝑡 = 𝑧 → ∀𝑦(𝑦 = 𝑡 → 𝜒))))
7	6	albidv 1953	. 2 ⊢ (𝜑 → (∀𝑡(𝑡 = 𝑧 → ∀𝑥(𝑥 = 𝑡 → 𝜓)) ↔ ∀𝑡(𝑡 = 𝑧 → ∀𝑦(𝑦 = 𝑡 → 𝜒))))
8		dfsb 2101	. 2 ⊢ ([𝑧 / 𝑥]𝜓 ↔ ∀𝑡(𝑡 = 𝑧 → ∀𝑥(𝑥 = 𝑡 → 𝜓)))
9		dfsb 2101	. 2 ⊢ ([𝑧 / 𝑦]𝜒 ↔ ∀𝑡(𝑡 = 𝑧 → ∀𝑦(𝑦 = 𝑡 → 𝜒)))
10	7, 8, 9	3bitr4g 317	1 ⊢ (𝜑 → ([𝑧 / 𝑥]𝜓 ↔ [𝑧 / 𝑦]𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  [wsb 2099

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100

This theorem is used by:  cbvabdavw  36827