| Metamath Proof Explorer |
< Previous
Next >
Related theorems GIF version |
| Description: The intersection of a non-empty class abstraction exists. |
| Ref | Expression |
|---|---|
| intexab | ⊢ (∃xφ ↔ ∩{x∣φ} ∈ V) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | abn0 1715 | . 2 ⊢ (¬ {x∣φ} = ∅ ↔ ∃xφ) | |
| 2 | intex 1986 | . 2 ⊢ (¬ {x∣φ} = ∅ ↔ ∩{x∣φ} ∈ V) | |
| 3 | 1, 2 | bitr3 153 | 1 ⊢ (∃xφ ↔ ∩{x∣φ} ∈ V) |
| Colors of variables: wff set class |
| Syntax hints: ¬ wn 1 ↔ wb 127 ∃wex 678 {cab 1090 = wceq 1091 ∈ wcel 1092 Vcvv 1348 ∅c0 1707 ∩cint 1965 |
| This theorem is referenced by: intexrab 1988 cfval 3701 cffnon 3702 |
| This theorem was proved from axioms: ax-1 3 ax-2 4 ax-3 5 ax-mp 6 ax-4 673 ax-5 674 ax-6 675 ax-7 676 ax-gen 677 ax-8 798 ax-9 799 ax-10 800 ax-11 801 ax-12 802 ax-13 804 ax-14 805 ax-16 922 ax-17 925 ax-ext 1074 ax-rep 1075 |
| This theorem depends on definitions: df-bi 128 df-or 197 df-an 198 df-ex 679 df-sb 853 df-clab 1093 df-cleq 1097 df-clel 1099 df-ral 1205 df-v 1349 df-dif 1489 df-in 1491 df-ss 1492 df-nul 1708 df-int 1966 |