Theorem eqtrb 32675

Description: A transposition of equality. (Contributed by Thierry Arnoux, 20-Aug-2023.)

Assertion

Ref	Expression
eqtrb	⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵 = 𝐶))

Proof of Theorem eqtrb

Step	Hyp	Ref	Expression
1		simpl 486	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) → 𝐴 = 𝐵)
2		eqtr2 2785	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) → 𝐵 = 𝐶)
3	1, 2	jca 519	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) → (𝐴 = 𝐵 ∧ 𝐵 = 𝐶))
4		simpl 486	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) → 𝐴 = 𝐵)
5		eqtr 2784	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) → 𝐴 = 𝐶)
6	4, 5	jca 519	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) → (𝐴 = 𝐵 ∧ 𝐴 = 𝐶))
7	3, 6	impbii 211	1 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵 = 𝐶))

Syntax hints:   ↔ wb 208   ∧ wa 399   = wceq 1562

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1817  ax-4 1831  ax-5 1932  ax-6 1989  ax-7 2030  ax-9 2154  ax-ext 2736

This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1802  df-cleq 2756

This theorem is referenced by: (None)