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| Description: Equality theorem for partial ordering predicate. (Contributed by NM, 27-Mar-1997.) | 
| Ref | Expression | 
|---|---|
| poeq2 | ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐴 ↔ 𝑅 Po 𝐵)) | 
| Step | Hyp | Ref | Expression | 
|---|---|---|---|
| 1 | eqimss2 4042 | . . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴) | |
| 2 | poss 5593 | . . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝑅 Po 𝐴 → 𝑅 Po 𝐵)) | |
| 3 | 1, 2 | syl 17 | . 2 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐴 → 𝑅 Po 𝐵)) | 
| 4 | eqimss 4041 | . . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵) | |
| 5 | poss 5593 | . . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Po 𝐵 → 𝑅 Po 𝐴)) | |
| 6 | 4, 5 | syl 17 | . 2 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐵 → 𝑅 Po 𝐴)) | 
| 7 | 3, 6 | impbid 212 | 1 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐴 ↔ 𝑅 Po 𝐵)) | 
| Colors of variables: wff setvar class | 
| Syntax hints: → wi 4 ↔ wb 206 = wceq 1539 ⊆ wss 3950 Po wpo 5589 | 
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1794 ax-4 1808 ax-5 1909 ax-6 1966 ax-7 2006 ax-9 2117 ax-ext 2707 | 
| This theorem depends on definitions: df-bi 207 df-an 396 df-ex 1779 df-cleq 2728 df-ral 3061 df-ss 3967 df-po 5591 | 
| This theorem is referenced by: poeq12d 5596 posn 5770 dfpo2 6315 frfi 9322 ipo0 44473 | 
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