Theorem eqrdav 2021

Description: Deduce equality of classes from an equivalence of membership that depends on the membership variable. (Contributed by NM, 7-Nov-2008.)

Hypotheses

Ref	Expression
eqrdav.1	⊢ ((φ ∧ x ∈ A) → x ∈ 𝐶)
eqrdav.2	⊢ ((φ ∧ x ∈ B) → x ∈ 𝐶)
eqrdav.3	⊢ ((φ ∧ x ∈ 𝐶) → (x ∈ A ↔ x ∈ B))

Assertion

Ref	Expression
eqrdav	⊢ (φ → A = B)

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

Allowed substitution hint:   𝐶(x)

Proof of Theorem eqrdav

Step	Hyp	Ref	Expression
1		eqrdav.1	. . . 4 ⊢ ((φ ∧ x ∈ A) → x ∈ 𝐶)
2		eqrdav.3	. . . . . 6 ⊢ ((φ ∧ x ∈ 𝐶) → (x ∈ A ↔ x ∈ B))
3	2	biimpd 132	. . . . 5 ⊢ ((φ ∧ x ∈ 𝐶) → (x ∈ A → x ∈ B))
4	3	impancom 247	. . . 4 ⊢ ((φ ∧ x ∈ A) → (x ∈ 𝐶 → x ∈ B))
5	1, 4	mpd 13	. . 3 ⊢ ((φ ∧ x ∈ A) → x ∈ B)
6		eqrdav.2	. . . 4 ⊢ ((φ ∧ x ∈ B) → x ∈ 𝐶)
7	2	exbiri 364	. . . . . 6 ⊢ (φ → (x ∈ 𝐶 → (x ∈ B → x ∈ A)))
8	7	com23 72	. . . . 5 ⊢ (φ → (x ∈ B → (x ∈ 𝐶 → x ∈ A)))
9	8	imp 115	. . . 4 ⊢ ((φ ∧ x ∈ B) → (x ∈ 𝐶 → x ∈ A))
10	6, 9	mpd 13	. . 3 ⊢ ((φ ∧ x ∈ B) → x ∈ A)
11	5, 10	impbida 515	. 2 ⊢ (φ → (x ∈ A ↔ x ∈ B))
12	11	eqrdv 2020	1 ⊢ (φ → A = B)

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1228   ∈ wcel 1374

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1316  ax-gen 1318  ax-17 1400  ax-ext 2004

This theorem depends on definitions:  df-bi 110  df-cleq 2015

This theorem is referenced by: (None)