Theorem eqrdav 2759

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	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 ∈ 𝐶)
2		eqrdav.2	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐵) → 𝑥 ∈ 𝐶)
3		eqrdav.3	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐶) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
4	1, 2, 3	bibiad 853	. 2 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
5	4	eqrdv 2758	1 ⊢ (𝜑 → 𝐴 = 𝐵)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569   ∈ wcel 2144

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-cleq 2752

This theorem is used by:  boxcutc  8970  supminf  13063  f1omvdconj  19661  fmucndlem  24610  lsmsnorb  33989  ballotlemsima  35191  supminfxr  46550