Theorem eqelb 38608

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 483	. . . 4 ⊢ ((𝐵 = 𝐴 ∧ 𝐴 ∈ 𝐶) → 𝐵 = 𝐴)
2		eqeltr 38607	. . . 4 ⊢ ((𝐵 = 𝐴 ∧ 𝐴 ∈ 𝐶) → 𝐵 ∈ 𝐶)
3	1, 2	jca 516	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐴 ∈ 𝐶) → (𝐵 = 𝐴 ∧ 𝐵 ∈ 𝐶))
4		eqcom 2746	. . . 4 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
5	4	anbi1i 630	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐴 ∈ 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶))
6	4	anbi1i 630	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐵 ∈ 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶))
7	3, 5, 6	3imtr3i 292	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶) → (𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶))
8		simpl 483	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶) → 𝐴 = 𝐵)
9		eqeltr 38607	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶) → 𝐴 ∈ 𝐶)
10	8, 9	jca 516	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶) → (𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶))
11	7, 10	impbii 210	1 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 ∈ 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵 ∈ 𝐶))

Syntax hints:   ↔ wb 207   ∧ wa 396   = wceq 1547   ∈ wcel 2119

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711

This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1787  df-cleq 2731  df-clel 2814

This theorem is referenced by:  eldmressnALTV  38646  raldmqseu  38732  inxpxrn  38785