Theorem eqelb 39439

Description: Substitution of equal classes into element relation. (Contributed by Peter Mazsa, 17-Jul-2019.)

Assertion

Ref	Expression
eqelb	⊢ ((𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶))

Proof of Theorem eqelb

Step	Hyp	Ref	Expression
1		simpl 488	. . . 4 ⊢ ((𝐵 = 𝐴 ∧ 𝐴 ∈ 𝐶) → 𝐵 = 𝐴)
2		eqeltr 39438	. . . 4 ⊢ ((𝐵 = 𝐴 ∧ 𝐴 ∈ 𝐶) → 𝐵 ∈ 𝐶)
3	1, 2	jca 521	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐴 ∈ 𝐶) → (𝐵 = 𝐴 ∧ 𝐵 ∈ 𝐶))
4		eqcom 2770	. . . 4 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
5	4	anbi1i 636	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐴 ∈ 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶))
6	4	anbi1i 636	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐵 ∈ 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶))
7	3, 5, 6	3imtr3i 294	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶) → (𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶))
8		simpl 488	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶) → 𝐴 = 𝐵)
9		eqeltr 39438	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶) → 𝐴 ∈ 𝐶)
10	8, 9	jca 521	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶) → (𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶))
11	7, 10	impbii 212	1 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2147

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2755  df-clel 2839

This theorem is used by:  eldmressnALTV  39477  raldmqseu  39563  inxpxrn  39616