Theorem ceqsralbv 3615

Description: Elimination of a restricted universal quantifier, using implicit substitution. (Contributed by Scott Fenton, 7-Dec-2020.)

Hypothesis

Ref	Expression
ceqsrexv.1	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
ceqsralbv	⊢ (∀𝑥 ∈ 𝐵 (𝑥 = 𝐴 → 𝜑) ↔ (𝐴 ∈ 𝐵 → 𝜓))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜓,𝑥

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem ceqsralbv

Step	Hyp	Ref	Expression
1		ceqsrexv.1	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
2	1	notbid 321	. . . 4 ⊢ (𝑥 = 𝐴 → (¬ 𝜑 ↔ ¬ 𝜓))
3	2	ceqsrexbv 3614	. . 3 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ ¬ 𝜑) ↔ (𝐴 ∈ 𝐵 ∧ ¬ 𝜓))
4		rexanali 3118	. . 3 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ ¬ 𝜑) ↔ ¬ ∀𝑥 ∈ 𝐵 (𝑥 = 𝐴 → 𝜑))
5		annim 408	. . 3 ⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝜓) ↔ ¬ (𝐴 ∈ 𝐵 → 𝜓))
6	3, 4, 5	3bitr3i 304	. 2 ⊢ (¬ ∀𝑥 ∈ 𝐵 (𝑥 = 𝐴 → 𝜑) ↔ ¬ (𝐴 ∈ 𝐵 → 𝜓))
7	6	con4bii 324	1 ⊢ (∀𝑥 ∈ 𝐵 (𝑥 = 𝐴 → 𝜑) ↔ (𝐴 ∈ 𝐵 → 𝜓))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1569   ∈ wcel 2142  ∀wral 3078  ∃wrex 3088

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3079  df-rex 3089

This theorem is used by:  ref5  38996