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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 2082 | . . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∧ ¬ 𝜓)) | |
2 | sban 1884 | . . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ ¬ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥] ¬ 𝜓)) | |
3 | df-clab 2082 | . . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑) | |
4 | 3 | bicomi 131 | . . . 4 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}) |
5 | sbn 1881 | . . . . 5 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ [𝑦 / 𝑥]𝜓) | |
6 | df-clab 2082 | . . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓) | |
7 | 5, 6 | xchbinxr 646 | . . . 4 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓}) |
8 | 4, 7 | anbi12i 449 | . . 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 3135 | 1 ⊢ ({𝑥 ∣ 𝜑} ∖ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)} |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 ∧ wa 103 = wceq 1296 ∈ wcel 1445 [wsb 1699 {cab 2081 ∖ cdif 3010 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 582 ax-in2 583 ax-io 668 ax-5 1388 ax-7 1389 ax-gen 1390 ax-ie1 1434 ax-ie2 1435 ax-8 1447 ax-10 1448 ax-11 1449 ax-i12 1450 ax-bndl 1451 ax-4 1452 ax-17 1471 ax-i9 1475 ax-ial 1479 ax-i5r 1480 ax-ext 2077 |
This theorem depends on definitions: df-bi 116 df-tru 1299 df-fal 1302 df-nf 1402 df-sb 1700 df-clab 2082 df-cleq 2088 df-clel 2091 df-nfc 2224 df-v 2635 df-dif 3015 |
This theorem is referenced by: notab 3285 difrab 3289 notrab 3292 imadiflem 5127 imadif 5128 |
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