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Mirrors > Home > ILE Home > Th. List > releq | GIF version |
Description: Equality theorem for the relation predicate. (Contributed by NM, 1-Aug-1994.) |
Ref | Expression |
---|---|
releq | ⊢ (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | sseq1 3178 | . 2 ⊢ (𝐴 = 𝐵 → (𝐴 ⊆ (V × V) ↔ 𝐵 ⊆ (V × V))) | |
2 | df-rel 4630 | . 2 ⊢ (Rel 𝐴 ↔ 𝐴 ⊆ (V × V)) | |
3 | df-rel 4630 | . 2 ⊢ (Rel 𝐵 ↔ 𝐵 ⊆ (V × V)) | |
4 | 1, 2, 3 | 3bitr4g 223 | 1 ⊢ (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 105 = wceq 1353 Vcvv 2737 ⊆ wss 3129 × cxp 4621 Rel wrel 4628 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-5 1447 ax-7 1448 ax-gen 1449 ax-ie1 1493 ax-ie2 1494 ax-8 1504 ax-11 1506 ax-4 1510 ax-17 1526 ax-i9 1530 ax-ial 1534 ax-i5r 1535 ax-ext 2159 |
This theorem depends on definitions: df-bi 117 df-nf 1461 df-sb 1763 df-clab 2164 df-cleq 2170 df-clel 2173 df-in 3135 df-ss 3142 df-rel 4630 |
This theorem is referenced by: releqi 4706 releqd 4707 dfrel2 5075 tposfn2 6261 ereq1 6536 |
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