Theorem alexeqg 3605

Description: Two ways to express substitution of 𝐴 for 𝑥 in 𝜑. This is the analogue for classes of sbalex 2279. (Contributed by NM, 2-Mar-1995.) (Revised by BJ, 27-Apr-2019.)

Assertion

Ref	Expression
alexeqg	⊢ (𝐴 ∈ 𝑉 → (∀𝑥(𝑥 = 𝐴 → 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))

Distinct variable group:   𝑥,𝐴

Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)

Proof of Theorem alexeqg

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eqeq2 2773	. . . . 5 ⊢ (𝑦 = 𝐴 → (𝑥 = 𝑦 ↔ 𝑥 = 𝐴))
2	1	anbi1d 643	. . . 4 ⊢ (𝑦 = 𝐴 → ((𝑥 = 𝑦 ∧ 𝜑) ↔ (𝑥 = 𝐴 ∧ 𝜑)))
3	2	exbidv 1954	. . 3 ⊢ (𝑦 = 𝐴 → (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
4	1	imbi1d 344	. . . 4 ⊢ (𝑦 = 𝐴 → ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝐴 → 𝜑)))
5	4	albidv 1953	. . 3 ⊢ (𝑦 = 𝐴 → (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝐴 → 𝜑)))
6		sbalex 2279	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
7	3, 5, 6	vtoclbg 3520	. 2 ⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝐴 → 𝜑)))
8	7	bicomd 226	1 ⊢ (𝐴 ∈ 𝑉 → (∀𝑥(𝑥 = 𝐴 → 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   = wceq 1570  ∃wex 1812   ∈ wcel 2145

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  ax-8 2147  ax-9 2155  ax-10 2178  ax-12 2213  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836

This theorem is used by:  ceqex  3606