| Mathbox for Thierry Arnoux |
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| Mirrors > Home > MPE Home > Th. List > Mathboxes > eqtrb | Structured version Visualization version GIF version | ||
| Description: A transposition of equality. (Contributed by Thierry Arnoux, 20-Aug-2023.) |
| Ref | Expression |
|---|---|
| eqtrb | ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵 = 𝐶)) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | simpl 482 | . . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) → 𝐴 = 𝐵) | |
| 2 | eqtr2 2761 | . . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) → 𝐵 = 𝐶) | |
| 3 | 1, 2 | jca 511 | . 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) → (𝐴 = 𝐵 ∧ 𝐵 = 𝐶)) |
| 4 | simpl 482 | . . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) → 𝐴 = 𝐵) | |
| 5 | eqtr 2760 | . . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) → 𝐴 = 𝐶) | |
| 6 | 4, 5 | jca 511 | . 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) → (𝐴 = 𝐵 ∧ 𝐴 = 𝐶)) |
| 7 | 3, 6 | impbii 209 | 1 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵 = 𝐶)) |
| Colors of variables: wff setvar class |
| Syntax hints: ↔ wb 206 ∧ wa 395 = wceq 1540 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2007 ax-9 2118 ax-ext 2708 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-ex 1780 df-cleq 2729 |
| This theorem is referenced by: (None) |
| Copyright terms: Public domain | W3C validator |