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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 3170 | . 2 ⊢ (𝐴 = 𝐵 → (𝐴 ⊆ (V × V) ↔ 𝐵 ⊆ (V × V))) | |
2 | df-rel 4618 | . 2 ⊢ (Rel 𝐴 ↔ 𝐴 ⊆ (V × V)) | |
3 | df-rel 4618 | . 2 ⊢ (Rel 𝐵 ↔ 𝐵 ⊆ (V × V)) | |
4 | 1, 2, 3 | 3bitr4g 222 | 1 ⊢ (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 104 = wceq 1348 Vcvv 2730 ⊆ wss 3121 × cxp 4609 Rel wrel 4616 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-5 1440 ax-7 1441 ax-gen 1442 ax-ie1 1486 ax-ie2 1487 ax-8 1497 ax-11 1499 ax-4 1503 ax-17 1519 ax-i9 1523 ax-ial 1527 ax-i5r 1528 ax-ext 2152 |
This theorem depends on definitions: df-bi 116 df-nf 1454 df-sb 1756 df-clab 2157 df-cleq 2163 df-clel 2166 df-in 3127 df-ss 3134 df-rel 4618 |
This theorem is referenced by: releqi 4694 releqd 4695 dfrel2 5061 tposfn2 6245 ereq1 6520 |
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