Theorem eqtrb 33070

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 488	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) → 𝐴 = 𝐵)
2		eqtr2 2781	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) → 𝐵 = 𝐶)
3	1, 2	jca 521	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) → (𝐴 = 𝐵 ∧ 𝐵 = 𝐶))
4		simpl 488	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) → 𝐴 = 𝐵)
5		eqtr 2780	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) → 𝐴 = 𝐶)
6	4, 5	jca 521	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) → (𝐴 = 𝐵 ∧ 𝐴 = 𝐶))
7	3, 6	impbii 212	1 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵 = 𝐶))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1569

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-cleq 2752

This theorem is used by: (None)