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Mirrors > Home > ILE Home > Th. List > fneq2d | GIF version |
Description: Equality deduction for function predicate with domain. (Contributed by Paul Chapman, 22-Jun-2011.) |
Ref | Expression |
---|---|
fneq2d.1 | ⊢ (𝜑 → 𝐴 = 𝐵) |
Ref | Expression |
---|---|
fneq2d | ⊢ (𝜑 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | fneq2d.1 | . 2 ⊢ (𝜑 → 𝐴 = 𝐵) | |
2 | fneq2 5262 | . 2 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵)) | |
3 | 1, 2 | syl 14 | 1 ⊢ (𝜑 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 104 = wceq 1335 Fn wfn 5168 |
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 1427 ax-gen 1429 ax-4 1490 ax-17 1506 ax-ext 2139 |
This theorem depends on definitions: df-bi 116 df-cleq 2150 df-fn 5176 |
This theorem is referenced by: fneq12d 5265 acfun 7145 ccfunen 7187 seq3shft 10750 |
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