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Mirrors > Home > MPE Home > Th. List > difab | Structured version Visualization version 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 2800 | . . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∧ ¬ 𝜓)) | |
2 | sban 2086 | . . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ ¬ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥] ¬ 𝜓)) | |
3 | df-clab 2800 | . . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑) | |
4 | 3 | bicomi 226 | . . . 4 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}) |
5 | sbn 2287 | . . . . 5 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ [𝑦 / 𝑥]𝜓) | |
6 | df-clab 2800 | . . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓) | |
7 | 5, 6 | xchbinxr 337 | . . . 4 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓}) |
8 | 4, 7 | anbi12i 628 | . . 3 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥] ¬ 𝜓) ↔ (𝑦 ∈ {𝑥 ∣ 𝜑} ∧ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓})) |
9 | 1, 2, 8 | 3bitrri 300 | . 2 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∧ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ 𝑦 ∈ {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)}) |
10 | 9 | difeqri 4101 | 1 ⊢ ({𝑥 ∣ 𝜑} ∖ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)} |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 ∧ wa 398 = wceq 1537 [wsb 2069 ∈ wcel 2114 {cab 2799 ∖ cdif 3933 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2116 ax-9 2124 ax-10 2145 ax-11 2161 ax-12 2177 ax-ext 2793 |
This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-tru 1540 df-ex 1781 df-nf 1785 df-sb 2070 df-clab 2800 df-cleq 2814 df-clel 2893 df-nfc 2963 df-v 3496 df-dif 3939 |
This theorem is referenced by: notab 4273 difrab 4277 notrab 4280 |
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