Theorem eqrdav 2080

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 142	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐶) → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
4	3	impancom 256	. . . 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 374	. . . . . 6 ⊢ (𝜑 → (𝑥 ∈ 𝐶 → (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴)))
8	7	com23 77	. . . . 5 ⊢ (𝜑 → (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐶 → 𝑥 ∈ 𝐴)))
9	8	imp 122	. . . 4 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐵) → (𝑥 ∈ 𝐶 → 𝑥 ∈ 𝐴))
10	6, 9	mpd 13	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐵) → 𝑥 ∈ 𝐴)
11	5, 10	impbida 560	. 2 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
12	11	eqrdv 2079	1 ⊢ (𝜑 → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103   = wceq 1284   ∈ wcel 1433

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-17 1459  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-cleq 2074

This theorem is referenced by:  supminfex  8685  fzdifsuc  9098