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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 2127 | . . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∧ ¬ 𝜓)) | |
2 | sban 1929 | . . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ ¬ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥] ¬ 𝜓)) | |
3 | df-clab 2127 | . . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑) | |
4 | 3 | bicomi 131 | . . . 4 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}) |
5 | sbn 1926 | . . . . 5 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ [𝑦 / 𝑥]𝜓) | |
6 | df-clab 2127 | . . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓) | |
7 | 5, 6 | xchbinxr 673 | . . . 4 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓}) |
8 | 4, 7 | anbi12i 456 | . . 3 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥] ¬ 𝜓) ↔ (𝑦 ∈ {𝑥 ∣ 𝜑} ∧ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓})) |
9 | 1, 2, 8 | 3bitrri 206 | . 2 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∧ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ 𝑦 ∈ {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)}) |
10 | 9 | difeqri 3201 | 1 ⊢ ({𝑥 ∣ 𝜑} ∖ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)} |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 ∧ wa 103 = wceq 1332 ∈ wcel 1481 [wsb 1736 {cab 2126 ∖ cdif 3073 |
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-in1 604 ax-in2 605 ax-io 699 ax-5 1424 ax-7 1425 ax-gen 1426 ax-ie1 1470 ax-ie2 1471 ax-8 1483 ax-10 1484 ax-11 1485 ax-i12 1486 ax-bndl 1487 ax-4 1488 ax-17 1507 ax-i9 1511 ax-ial 1515 ax-i5r 1516 ax-ext 2122 |
This theorem depends on definitions: df-bi 116 df-tru 1335 df-fal 1338 df-nf 1438 df-sb 1737 df-clab 2127 df-cleq 2133 df-clel 2136 df-nfc 2271 df-v 2691 df-dif 3078 |
This theorem is referenced by: notab 3351 difrab 3355 notrab 3358 imadiflem 5210 imadif 5211 |
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