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Mirrors > Home > MPE Home > Th. List > rusgrnumwwlkslem | Structured version Visualization version GIF version |
Description: Lemma for rusgrnumwwlks 28030. (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 6705 | . . . 4 ⊢ (𝑤 = 𝑌 → (𝑤‘0) = (𝑌‘0)) | |
2 | 1 | eqeq1d 2736 | . . 3 ⊢ (𝑤 = 𝑌 → ((𝑤‘0) = 𝑃 ↔ (𝑌‘0) = 𝑃)) |
3 | 2 | elrab 3595 | . 2 ⊢ (𝑌 ∈ {𝑤 ∈ 𝑍 ∣ (𝑤‘0) = 𝑃} ↔ (𝑌 ∈ 𝑍 ∧ (𝑌‘0) = 𝑃)) |
4 | ibar 532 | . . . . 5 ⊢ ((𝑌‘0) = 𝑃 → ((𝜑 ∧ 𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑 ∧ 𝜓)))) | |
5 | 3anass 1097 | . . . . . 6 ⊢ (((𝑌‘0) = 𝑃 ∧ 𝜑 ∧ 𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑 ∧ 𝜓))) | |
6 | 3ancoma 1100 | . . . . . 6 ⊢ (((𝑌‘0) = 𝑃 ∧ 𝜑 ∧ 𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)) | |
7 | 5, 6 | bitr3i 280 | . . . . 5 ⊢ (((𝑌‘0) = 𝑃 ∧ (𝜑 ∧ 𝜓)) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)) |
8 | 4, 7 | bitrdi 290 | . . . 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 3381 | . 2 ⊢ ((𝑌 ∈ 𝑍 ∧ (𝑌‘0) = 𝑃) → {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ 𝜓)} = {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)}) |
11 | 3, 10 | sylbi 220 | 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 209 ∧ wa 399 ∧ w3a 1089 = wceq 1543 ∈ wcel 2110 {crab 3058 ‘cfv 6369 0cc0 10712 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1803 ax-4 1817 ax-5 1918 ax-6 1976 ax-7 2016 ax-8 2112 ax-9 2120 ax-ext 2706 |
This theorem depends on definitions: df-bi 210 df-an 400 df-3an 1091 df-tru 1546 df-ex 1788 df-sb 2071 df-clab 2713 df-cleq 2726 df-clel 2812 df-rab 3063 df-v 3403 df-in 3864 df-ss 3874 df-uni 4810 df-br 5044 df-iota 6327 df-fv 6377 |
This theorem is referenced by: rusgrnumwwlks 28030 |
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