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Theorem rusgranumwlklem0 26475
 Description: Lemma 0 for rusgranumwlk 26484. (Contributed by Alexander van der Vekens, 23-Aug-2018.)
Assertion
Ref Expression
rusgranumwlklem0 (𝑌 ∈ {𝑤𝑍 ∣ (𝑤‘0) = 𝑃} → {𝑤𝑋 ∣ (𝜑𝜓)} = {𝑤𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)})
Distinct variable groups:   𝑤,𝑃   𝑤,𝑌   𝑤,𝑍
Allowed substitution hints:   𝜑(𝑤)   𝜓(𝑤)   𝑋(𝑤)

Proof of Theorem rusgranumwlklem0
StepHypRef Expression
1 fveq1 6102 . . . 4 (𝑤 = 𝑌 → (𝑤‘0) = (𝑌‘0))
21eqeq1d 2612 . . 3 (𝑤 = 𝑌 → ((𝑤‘0) = 𝑃 ↔ (𝑌‘0) = 𝑃))
32elrab 3331 . 2 (𝑌 ∈ {𝑤𝑍 ∣ (𝑤‘0) = 𝑃} ↔ (𝑌𝑍 ∧ (𝑌‘0) = 𝑃))
4 ibar 524 . . . . 5 ((𝑌‘0) = 𝑃 → ((𝜑𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑𝜓))))
5 3anass 1035 . . . . . 6 (((𝑌‘0) = 𝑃𝜑𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑𝜓)))
6 3ancoma 1038 . . . . . 6 (((𝑌‘0) = 𝑃𝜑𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓))
75, 6bitr3i 265 . . . . 5 (((𝑌‘0) = 𝑃 ∧ (𝜑𝜓)) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓))
84, 7syl6bb 275 . . . 4 ((𝑌‘0) = 𝑃 → ((𝜑𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)))
98ad2antlr 759 . . 3 (((𝑌𝑍 ∧ (𝑌‘0) = 𝑃) ∧ 𝑤𝑋) → ((𝜑𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)))
109rabbidva 3163 . 2 ((𝑌𝑍 ∧ (𝑌‘0) = 𝑃) → {𝑤𝑋 ∣ (𝜑𝜓)} = {𝑤𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)})
113, 10sylbi 206 1 (𝑌 ∈ {𝑤𝑍 ∣ (𝑤‘0) = 𝑃} → {𝑤𝑋 ∣ (𝜑𝜓)} = {𝑤𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)})
 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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