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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 5878 | . . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐴) = (𝐵 ∘ 𝐵)) |
3 | 2, 1 | sseq12d 4029 | . 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∘ 𝐴) ⊆ 𝐴 ↔ (𝐵 ∘ 𝐵) ⊆ 𝐵)) |
4 | 3 | cleq2lem 43598 | 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 1537 ⊆ wss 3963 ∘ ccom 5693 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1908 ax-6 1965 ax-7 2005 ax-8 2108 ax-9 2116 ax-ext 2706 |
This theorem depends on definitions: df-bi 207 df-an 396 df-ex 1777 df-sb 2063 df-clab 2713 df-cleq 2727 df-clel 2814 df-ss 3980 df-br 5149 df-opab 5211 df-co 5698 |
This theorem is referenced by: (None) |
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