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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 5208 | . . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐴) = (𝐵 ∘ 𝐵)) |
3 | 2, 1 | sseq12d 3597 | . 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∘ 𝐴) ⊆ 𝐴 ↔ (𝐵 ∘ 𝐵) ⊆ 𝐵)) |
4 | 3 | cleq2lem 36933 | 1 ⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑅 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 195 ∧ wa 383 = wceq 1475 ⊆ wss 3540 ∘ ccom 5042 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 |
This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-in 3547 df-ss 3554 df-br 4584 df-opab 4644 df-co 5047 |
This theorem is referenced by: (None) |
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