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Mirrors > Home > ILE Home > Th. List > difab | GIF version |
Description: Difference of two class abstractions. (Contributed by NM, 23-Oct-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) |
Ref | Expression |
---|---|
difab | ⊢ ({𝑥 ∣ 𝜑} ∖ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)} |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-clab 2068 | . . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∧ ¬ 𝜓)) | |
2 | sban 1870 | . . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ ¬ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥] ¬ 𝜓)) | |
3 | df-clab 2068 | . . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑) | |
4 | 3 | bicomi 130 | . . . 4 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}) |
5 | sbn 1867 | . . . . 5 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ [𝑦 / 𝑥]𝜓) | |
6 | df-clab 2068 | . . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓) | |
7 | 5, 6 | xchbinxr 640 | . . . 4 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓}) |
8 | 4, 7 | anbi12i 447 | . . 3 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥] ¬ 𝜓) ↔ (𝑦 ∈ {𝑥 ∣ 𝜑} ∧ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓})) |
9 | 1, 2, 8 | 3bitrri 205 | . 2 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∧ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ 𝑦 ∈ {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)}) |
10 | 9 | difeqri 3092 | 1 ⊢ ({𝑥 ∣ 𝜑} ∖ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)} |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 ∧ wa 102 = wceq 1284 ∈ wcel 1433 [wsb 1685 {cab 2067 ∖ cdif 2970 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 104 ax-ia2 105 ax-ia3 106 ax-in1 576 ax-in2 577 ax-io 662 ax-5 1376 ax-7 1377 ax-gen 1378 ax-ie1 1422 ax-ie2 1423 ax-8 1435 ax-10 1436 ax-11 1437 ax-i12 1438 ax-bndl 1439 ax-4 1440 ax-17 1459 ax-i9 1463 ax-ial 1467 ax-i5r 1468 ax-ext 2063 |
This theorem depends on definitions: df-bi 115 df-tru 1287 df-fal 1290 df-nf 1390 df-sb 1686 df-clab 2068 df-cleq 2074 df-clel 2077 df-nfc 2208 df-v 2603 df-dif 2975 |
This theorem is referenced by: notab 3234 difrab 3238 notrab 3241 imadiflem 4998 imadif 4999 |
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