Theorem eqelb 37565

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 482	. . . 4 ⊢ ((𝐵 = 𝐴 ∧ 𝐴 ∈ 𝐶) → 𝐵 = 𝐴)
2		eqeltr 37564	. . . 4 ⊢ ((𝐵 = 𝐴 ∧ 𝐴 ∈ 𝐶) → 𝐵 ∈ 𝐶)
3	1, 2	jca 511	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐴 ∈ 𝐶) → (𝐵 = 𝐴 ∧ 𝐵 ∈ 𝐶))
4		eqcom 2738	. . . 4 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
5	4	anbi1i 623	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐴 ∈ 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶))
6	4	anbi1i 623	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐵 ∈ 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶))
7	3, 5, 6	3imtr3i 291	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶) → (𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶))
8		simpl 482	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶) → 𝐴 = 𝐵)
9		eqeltr 37564	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶) → 𝐴 ∈ 𝐶)
10	8, 9	jca 511	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶) → (𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶))
11	7, 10	impbii 208	1 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶))

Syntax hints:   ↔ wb 205   ∧ wa 395   = wceq 1540   ∈ wcel 2105

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1781  df-cleq 2723  df-clel 2809

This theorem is referenced by:  eldmressnALTV  37604  inxpxrn  37729