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| Mirrors > Home > MPE Home > Th. List > Mathboxes > trcleq2lemRP | Structured version Visualization version GIF version | ||
| Description: Equality implies bijection. (Contributed by RP, 5-May-2020.) (Proof modification is discouraged.) |
| Ref | Expression |
|---|---|
| trcleq2lemRP | ⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑅 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵))) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | id 22 | . . . 4 ⊢ (𝐴 = 𝐵 → 𝐴 = 𝐵) | |
| 2 | 1, 1 | coeq12d 5875 | . . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐴) = (𝐵 ∘ 𝐵)) |
| 3 | 2, 1 | sseq12d 4017 | . 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∘ 𝐴) ⊆ 𝐴 ↔ (𝐵 ∘ 𝐵) ⊆ 𝐵)) |
| 4 | 3 | cleq2lem 43621 | 1 ⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑅 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵))) |
| Colors of variables: wff setvar class |
| Syntax hints: → wi 4 ↔ wb 206 ∧ wa 395 = wceq 1540 ⊆ wss 3951 ∘ ccom 5689 |
| 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-8 2110 ax-9 2118 ax-ext 2708 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-ex 1780 df-sb 2065 df-clab 2715 df-cleq 2729 df-clel 2816 df-ss 3968 df-br 5144 df-opab 5206 df-co 5694 |
| This theorem is referenced by: (None) |
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