Theorem eqrdav 2352

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 198	. . . . 5 ⊢ ((φ ∧ x ∈ C) → (x ∈ A → x ∈ B))
4	3	impancom 427	. . . 4 ⊢ ((φ ∧ x ∈ A) → (x ∈ C → x ∈ B))
5	1, 4	mpd 14	. . 3 ⊢ ((φ ∧ x ∈ A) → x ∈ B)
6		eqrdav.2	. . . 4 ⊢ ((φ ∧ x ∈ B) → x ∈ C)
7	2	exbiri 605	. . . . . 6 ⊢ (φ → (x ∈ C → (x ∈ B → x ∈ A)))
8	7	com23 72	. . . . 5 ⊢ (φ → (x ∈ B → (x ∈ C → x ∈ A)))
9	8	imp 418	. . . 4 ⊢ ((φ ∧ x ∈ B) → (x ∈ C → x ∈ A))
10	6, 9	mpd 14	. . 3 ⊢ ((φ ∧ x ∈ B) → x ∈ A)
11	5, 10	impbida 805	. 2 ⊢ (φ → (x ∈ A ↔ x ∈ B))
12	11	eqrdv 2351	1 ⊢ (φ → A = B)

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358   = wceq 1642   ∈ wcel 1710

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-an 360  df-cleq 2346

This theorem is used by: (None)