Theorem cbvsbdavw 36193

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 2023	. . . . . . 7 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝑡 ↔ 𝑦 = 𝑡))
2	1	adantl 481	. . . . . 6 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝑥 = 𝑡 ↔ 𝑦 = 𝑡))
3		cbvsbdavw.1	. . . . . 6 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝜓 ↔ 𝜒))
4	2, 3	imbi12d 344	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → ((𝑥 = 𝑡 → 𝜓) ↔ (𝑦 = 𝑡 → 𝜒)))
5	4	cbvaldvaw 2036	. . . 4 ⊢ (𝜑 → (∀𝑥(𝑥 = 𝑡 → 𝜓) ↔ ∀𝑦(𝑦 = 𝑡 → 𝜒)))
6	5	imbi2d 340	. . 3 ⊢ (𝜑 → ((𝑡 = 𝑧 → ∀𝑥(𝑥 = 𝑡 → 𝜓)) ↔ (𝑡 = 𝑧 → ∀𝑦(𝑦 = 𝑡 → 𝜒))))
7	6	albidv 1919	. 2 ⊢ (𝜑 → (∀𝑡(𝑡 = 𝑧 → ∀𝑥(𝑥 = 𝑡 → 𝜓)) ↔ ∀𝑡(𝑡 = 𝑧 → ∀𝑦(𝑦 = 𝑡 → 𝜒))))
8		df-sb 2064	. 2 ⊢ ([𝑧 / 𝑥]𝜓 ↔ ∀𝑡(𝑡 = 𝑧 → ∀𝑥(𝑥 = 𝑡 → 𝜓)))
9		df-sb 2064	. 2 ⊢ ([𝑧 / 𝑦]𝜒 ↔ ∀𝑡(𝑡 = 𝑧 → ∀𝑦(𝑦 = 𝑡 → 𝜒)))
10	7, 8, 9	3bitr4g 314	1 ⊢ (𝜑 → ([𝑧 / 𝑥]𝜓 ↔ [𝑧 / 𝑦]𝜒))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1537  [wsb 2063

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1779  df-sb 2064

This theorem is referenced by:  cbvabdavw  36195