Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
||
Mirrors > Home > MPE Home > Th. List > seeq2 | Structured version Visualization version GIF version |
Description: Equality theorem for the set-like predicate. (Contributed by Mario Carneiro, 24-Jun-2015.) |
Ref | Expression |
---|---|
seeq2 | ⊢ (𝐴 = 𝐵 → (𝑅 Se 𝐴 ↔ 𝑅 Se 𝐵)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eqimss2 3621 | . . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴) | |
2 | sess2 5007 | . . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝑅 Se 𝐴 → 𝑅 Se 𝐵)) | |
3 | 1, 2 | syl 17 | . 2 ⊢ (𝐴 = 𝐵 → (𝑅 Se 𝐴 → 𝑅 Se 𝐵)) |
4 | eqimss 3620 | . . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵) | |
5 | sess2 5007 | . . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Se 𝐵 → 𝑅 Se 𝐴)) | |
6 | 4, 5 | syl 17 | . 2 ⊢ (𝐴 = 𝐵 → (𝑅 Se 𝐵 → 𝑅 Se 𝐴)) |
7 | 3, 6 | impbid 201 | 1 ⊢ (𝐴 = 𝐵 → (𝑅 Se 𝐴 ↔ 𝑅 Se 𝐵)) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 195 = wceq 1475 ⊆ wss 3540 Se wse 4995 |
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 ax-sep 4709 |
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-ral 2901 df-rab 2905 df-v 3175 df-in 3547 df-ss 3554 df-se 4998 |
This theorem is referenced by: oieq2 8301 |
Copyright terms: Public domain | W3C validator |