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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 2180 | . . 3 ⊢ (𝐴 = 𝐵 → (dom 𝑅 = 𝐴 ↔ dom 𝑅 = 𝐵)) | |
2 | 1 | 3anbi2d 1312 | . 2 ⊢ (𝐴 = 𝐵 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))) |
3 | df-er 6511 | . 2 ⊢ (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅)) | |
4 | df-er 6511 | . 2 ⊢ (𝑅 Er 𝐵 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅)) | |
5 | 2, 3, 4 | 3bitr4g 222 | 1 ⊢ (𝐴 = 𝐵 → (𝑅 Er 𝐴 ↔ 𝑅 Er 𝐵)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 104 ∧ w3a 973 = wceq 1348 ∪ cun 3119 ⊆ wss 3121 ◡ccnv 4608 dom cdm 4609 ∘ ccom 4613 Rel wrel 4614 Er wer 6508 |
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-gen 1442 ax-4 1503 ax-17 1519 ax-ext 2152 |
This theorem depends on definitions: df-bi 116 df-3an 975 df-cleq 2163 df-er 6511 |
This theorem is referenced by: iserd 6537 |
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