Theorem sbequbidv 36785

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

Hypotheses

Ref	Expression
sbequbidv.1	⊢ (𝜑 → 𝑢 = 𝑣)
sbequbidv.2	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
sbequbidv	⊢ (𝜑 → ([𝑢 / 𝑥]𝜓 ↔ [𝑣 / 𝑥]𝜒))

Distinct variable group:   𝜑,𝑥

Allowed substitution hints:   𝜑(𝑣, 𝑢)   𝜓(𝑥, 𝑣, 𝑢)   𝜒(𝑥, 𝑣, 𝑢)

Proof of Theorem sbequbidv

Dummy variable 𝑡 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbequbidv.1	. . . . 5 ⊢ (𝜑 → 𝑢 = 𝑣)
2		equequ2 2059	. . . . 5 ⊢ (𝑢 = 𝑣 → (𝑡 = 𝑢 ↔ 𝑡 = 𝑣))
3	1, 2	syl 18	. . . 4 ⊢ (𝜑 → (𝑡 = 𝑢 ↔ 𝑡 = 𝑣))
4		sbequbidv.2	. . . . . 6 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
5	4	imbi2d 343	. . . . 5 ⊢ (𝜑 → ((𝑥 = 𝑡 → 𝜓) ↔ (𝑥 = 𝑡 → 𝜒)))
6	5	albidv 1953	. . . 4 ⊢ (𝜑 → (∀𝑥(𝑥 = 𝑡 → 𝜓) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜒)))
7	3, 6	imbi12d 347	. . 3 ⊢ (𝜑 → ((𝑡 = 𝑢 → ∀𝑥(𝑥 = 𝑡 → 𝜓)) ↔ (𝑡 = 𝑣 → ∀𝑥(𝑥 = 𝑡 → 𝜒))))
8	7	albidv 1953	. 2 ⊢ (𝜑 → (∀𝑡(𝑡 = 𝑢 → ∀𝑥(𝑥 = 𝑡 → 𝜓)) ↔ ∀𝑡(𝑡 = 𝑣 → ∀𝑥(𝑥 = 𝑡 → 𝜒))))
9		dfsb 2101	. 2 ⊢ ([𝑢 / 𝑥]𝜓 ↔ ∀𝑡(𝑡 = 𝑢 → ∀𝑥(𝑥 = 𝑡 → 𝜓)))
10		dfsb 2101	. 2 ⊢ ([𝑣 / 𝑥]𝜒 ↔ ∀𝑡(𝑡 = 𝑣 → ∀𝑥(𝑥 = 𝑡 → 𝜒)))
11	8, 9, 10	3bitr4g 317	1 ⊢ (𝜑 → ([𝑢 / 𝑥]𝜓 ↔ [𝑣 / 𝑥]𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀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: (None)