Theorem eqrdav 2737

Description: Deduce equality of classes from an equivalence of membership that depends on the membership variable. (Contributed by NM, 7-Nov-2008.) (Proof shortened by Wolf Lammen, 19-Nov-2019.)

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 228	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐶) → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
4	3	impancom 451	. . . 4 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝑥 ∈ 𝐶 → 𝑥 ∈ 𝐵))
5	1, 4	mpd 15	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 ∈ 𝐵)
6		eqrdav.2	. . . 4 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐵) → 𝑥 ∈ 𝐶)
7	2	biimprd 247	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐶) → (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴))
8	7	impancom 451	. . . 4 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐵) → (𝑥 ∈ 𝐶 → 𝑥 ∈ 𝐴))
9	6, 8	mpd 15	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐵) → 𝑥 ∈ 𝐴)
10	5, 9	impbida 797	. 2 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
11	10	eqrdv 2736	1 ⊢ (𝜑 → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1539   ∈ wcel 2108

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1784  df-cleq 2730

This theorem is referenced by:  boxcutc  8687  supminf  12604  f1omvdconj  18969  fmucndlem  23351  lsmsnorb  31481  ballotlemsima  32382  supminfxr  42894