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| Description: Lemma for rusgrnumwwlks 29994. (Contributed by Alexander van der Vekens, 23-Aug-2018.) | 
| Ref | Expression | 
|---|---|
| rusgrnumwwlkslem | ⊢ (𝑌 ∈ {𝑤 ∈ 𝑍 ∣ (𝑤‘0) = 𝑃} → {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ 𝜓)} = {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)}) | 
| Step | Hyp | Ref | Expression | 
|---|---|---|---|
| 1 | fveq1 6905 | . . . 4 ⊢ (𝑤 = 𝑌 → (𝑤‘0) = (𝑌‘0)) | |
| 2 | 1 | eqeq1d 2739 | . . 3 ⊢ (𝑤 = 𝑌 → ((𝑤‘0) = 𝑃 ↔ (𝑌‘0) = 𝑃)) | 
| 3 | 2 | elrab 3692 | . 2 ⊢ (𝑌 ∈ {𝑤 ∈ 𝑍 ∣ (𝑤‘0) = 𝑃} ↔ (𝑌 ∈ 𝑍 ∧ (𝑌‘0) = 𝑃)) | 
| 4 | ibar 528 | . . . . 5 ⊢ ((𝑌‘0) = 𝑃 → ((𝜑 ∧ 𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑 ∧ 𝜓)))) | |
| 5 | 3anass 1095 | . . . . . 6 ⊢ (((𝑌‘0) = 𝑃 ∧ 𝜑 ∧ 𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑 ∧ 𝜓))) | |
| 6 | 3ancoma 1098 | . . . . . 6 ⊢ (((𝑌‘0) = 𝑃 ∧ 𝜑 ∧ 𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)) | |
| 7 | 5, 6 | bitr3i 277 | . . . . 5 ⊢ (((𝑌‘0) = 𝑃 ∧ (𝜑 ∧ 𝜓)) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)) | 
| 8 | 4, 7 | bitrdi 287 | . . . 4 ⊢ ((𝑌‘0) = 𝑃 → ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓))) | 
| 9 | 8 | ad2antlr 727 | . . 3 ⊢ (((𝑌 ∈ 𝑍 ∧ (𝑌‘0) = 𝑃) ∧ 𝑤 ∈ 𝑋) → ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓))) | 
| 10 | 9 | rabbidva 3443 | . 2 ⊢ ((𝑌 ∈ 𝑍 ∧ (𝑌‘0) = 𝑃) → {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ 𝜓)} = {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)}) | 
| 11 | 3, 10 | sylbi 217 | 1 ⊢ (𝑌 ∈ {𝑤 ∈ 𝑍 ∣ (𝑤‘0) = 𝑃} → {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ 𝜓)} = {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)}) | 
| Colors of variables: wff setvar class | 
| Syntax hints: → wi 4 ↔ wb 206 ∧ wa 395 ∧ w3a 1087 = wceq 1540 ∈ wcel 2108 {crab 3436 ‘cfv 6561 0cc0 11155 | 
| 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-3an 1089 df-tru 1543 df-ex 1780 df-sb 2065 df-clab 2715 df-cleq 2729 df-clel 2816 df-rab 3437 df-v 3482 df-ss 3968 df-uni 4908 df-br 5144 df-iota 6514 df-fv 6569 | 
| This theorem is referenced by: rusgrnumwwlks 29994 | 
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