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Mirrors > Home > ILE Home > Th. List > fneq2 | GIF version |
Description: Equality theorem for function predicate with domain. (Contributed by NM, 1-Aug-1994.) |
Ref | Expression |
---|---|
fneq2 | ⊢ (A = B → (𝐹 Fn A ↔ 𝐹 Fn B)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eqeq2 2046 | . . 3 ⊢ (A = B → (dom 𝐹 = A ↔ dom 𝐹 = B)) | |
2 | 1 | anbi2d 437 | . 2 ⊢ (A = B → ((Fun 𝐹 ∧ dom 𝐹 = A) ↔ (Fun 𝐹 ∧ dom 𝐹 = B))) |
3 | df-fn 4848 | . 2 ⊢ (𝐹 Fn A ↔ (Fun 𝐹 ∧ dom 𝐹 = A)) | |
4 | df-fn 4848 | . 2 ⊢ (𝐹 Fn B ↔ (Fun 𝐹 ∧ dom 𝐹 = B)) | |
5 | 2, 3, 4 | 3bitr4g 212 | 1 ⊢ (A = B → (𝐹 Fn A ↔ 𝐹 Fn B)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∧ wa 97 ↔ wb 98 = wceq 1242 dom cdm 4288 Fun wfun 4839 Fn wfn 4840 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-5 1333 ax-gen 1335 ax-4 1397 ax-17 1416 ax-ext 2019 |
This theorem depends on definitions: df-bi 110 df-cleq 2030 df-fn 4848 |
This theorem is referenced by: fneq2d 4933 fneq2i 4937 feq2 4974 foeq2 5046 f1o00 5104 eqfnfv2 5209 tfr0 5878 tfrlemisucaccv 5880 tfrlemi1 5887 tfrlemi14d 5888 tfrexlem 5889 0fz1 8679 |
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