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Mirrors > Home > MPE Home > Th. List > rusgranumwlklem0 | Structured version Visualization version GIF version |
Description: Lemma 0 for rusgranumwlk 26484. (Contributed by Alexander van der Vekens, 23-Aug-2018.) |
Ref | Expression |
---|---|
rusgranumwlklem0 | ⊢ (𝑌 ∈ {𝑤 ∈ 𝑍 ∣ (𝑤‘0) = 𝑃} → {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ 𝜓)} = {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)}) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | fveq1 6102 | . . . 4 ⊢ (𝑤 = 𝑌 → (𝑤‘0) = (𝑌‘0)) | |
2 | 1 | eqeq1d 2612 | . . 3 ⊢ (𝑤 = 𝑌 → ((𝑤‘0) = 𝑃 ↔ (𝑌‘0) = 𝑃)) |
3 | 2 | elrab 3331 | . 2 ⊢ (𝑌 ∈ {𝑤 ∈ 𝑍 ∣ (𝑤‘0) = 𝑃} ↔ (𝑌 ∈ 𝑍 ∧ (𝑌‘0) = 𝑃)) |
4 | ibar 524 | . . . . 5 ⊢ ((𝑌‘0) = 𝑃 → ((𝜑 ∧ 𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑 ∧ 𝜓)))) | |
5 | 3anass 1035 | . . . . . 6 ⊢ (((𝑌‘0) = 𝑃 ∧ 𝜑 ∧ 𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑 ∧ 𝜓))) | |
6 | 3ancoma 1038 | . . . . . 6 ⊢ (((𝑌‘0) = 𝑃 ∧ 𝜑 ∧ 𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)) | |
7 | 5, 6 | bitr3i 265 | . . . . 5 ⊢ (((𝑌‘0) = 𝑃 ∧ (𝜑 ∧ 𝜓)) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)) |
8 | 4, 7 | syl6bb 275 | . . . 4 ⊢ ((𝑌‘0) = 𝑃 → ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓))) |
9 | 8 | ad2antlr 759 | . . 3 ⊢ (((𝑌 ∈ 𝑍 ∧ (𝑌‘0) = 𝑃) ∧ 𝑤 ∈ 𝑋) → ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓))) |
10 | 9 | rabbidva 3163 | . 2 ⊢ ((𝑌 ∈ 𝑍 ∧ (𝑌‘0) = 𝑃) → {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ 𝜓)} = {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)}) |
11 | 3, 10 | sylbi 206 | 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 195 ∧ wa 383 ∧ w3a 1031 = wceq 1475 ∈ wcel 1977 {crab 2900 ‘cfv 5804 0cc0 9815 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 |
This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ral 2901 df-rex 2902 df-rab 2905 df-v 3175 df-uni 4373 df-br 4584 df-iota 5768 df-fv 5812 |
This theorem is referenced by: rusgranumwlks 26483 rusgrnumwwlks 41177 |
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