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| Mirrors > Home > MPE Home > Th. List > rabeq0w | Structured version Visualization version GIF version | ||
| Description: Condition for a restricted class abstraction to be empty. Version of rabeq0 4388 using implicit substitution, which does not require ax-10 2141, ax-11 2157, ax-12 2177, but requires ax-8 2110. (Contributed by GG, 30-Sep-2024.) | 
| Ref | Expression | 
|---|---|
| rabeq0w.1 | ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) | 
| Ref | Expression | 
|---|---|
| rabeq0w | ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} = ∅ ↔ ∀𝑦 ∈ 𝐴 ¬ 𝜓) | 
| Step | Hyp | Ref | Expression | 
|---|---|---|---|
| 1 | eleq1w 2824 | . . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴)) | |
| 2 | rabeq0w.1 | . . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) | |
| 3 | 1, 2 | anbi12d 632 | . . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (𝑦 ∈ 𝐴 ∧ 𝜓))) | 
| 4 | 3 | ab0w 4379 | . 2 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} = ∅ ↔ ∀𝑦 ¬ (𝑦 ∈ 𝐴 ∧ 𝜓)) | 
| 5 | df-rab 3437 | . . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} | |
| 6 | 5 | eqeq1i 2742 | . 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} = ∅) | 
| 7 | raln 3069 | . 2 ⊢ (∀𝑦 ∈ 𝐴 ¬ 𝜓 ↔ ∀𝑦 ¬ (𝑦 ∈ 𝐴 ∧ 𝜓)) | |
| 8 | 4, 6, 7 | 3bitr4i 303 | 1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} = ∅ ↔ ∀𝑦 ∈ 𝐴 ¬ 𝜓) | 
| Colors of variables: wff setvar class | 
| Syntax hints: ¬ wn 3 → wi 4 ↔ wb 206 ∧ wa 395 ∀wal 1538 = wceq 1540 ∈ wcel 2108 {cab 2714 ∀wral 3061 {crab 3436 ∅c0 4333 | 
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2007 ax-8 2110 ax-9 2118 ax-ext 2708 | 
| This theorem depends on definitions: df-bi 207 df-an 396 df-tru 1543 df-fal 1553 df-ex 1780 df-sb 2065 df-clab 2715 df-cleq 2729 df-clel 2816 df-ral 3062 df-rab 3437 df-dif 3954 df-nul 4334 | 
| This theorem is referenced by: dffr2 5646 frc 5648 frirr 5661 | 
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