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Mirrors > Home > ILE Home > Th. List > ereq2 | GIF version |
Description: Equality theorem for equivalence predicate. (Contributed by Mario Carneiro, 12-Aug-2015.) |
Ref | Expression |
---|---|
ereq2 | ⊢ (𝐴 = 𝐵 → (𝑅 Er 𝐴 ↔ 𝑅 Er 𝐵)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eqeq2 2199 | . . 3 ⊢ (𝐴 = 𝐵 → (dom 𝑅 = 𝐴 ↔ dom 𝑅 = 𝐵)) | |
2 | 1 | 3anbi2d 1328 | . 2 ⊢ (𝐴 = 𝐵 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))) |
3 | df-er 6558 | . 2 ⊢ (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅)) | |
4 | df-er 6558 | . 2 ⊢ (𝑅 Er 𝐵 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅)) | |
5 | 2, 3, 4 | 3bitr4g 223 | 1 ⊢ (𝐴 = 𝐵 → (𝑅 Er 𝐴 ↔ 𝑅 Er 𝐵)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 105 ∧ w3a 980 = wceq 1364 ∪ cun 3142 ⊆ wss 3144 ◡ccnv 4643 dom cdm 4644 ∘ ccom 4648 Rel wrel 4649 Er wer 6555 |
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 1458 ax-gen 1460 ax-4 1521 ax-17 1537 ax-ext 2171 |
This theorem depends on definitions: df-bi 117 df-3an 982 df-cleq 2182 df-er 6558 |
This theorem is referenced by: iserd 6584 |
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