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Theorem rusgrnumwwlkslem 27110
Description: Lemma for rusgrnumwwlks 27115. (Contributed by Alexander van der Vekens, 23-Aug-2018.)
Assertion
Ref Expression
rusgrnumwwlkslem (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 rusgrnumwwlkslem
StepHypRef Expression
1 fveq1 6403 . . . 4 (𝑤 = 𝑌 → (𝑤‘0) = (𝑌‘0))
21eqeq1d 2808 . . 3 (𝑤 = 𝑌 → ((𝑤‘0) = 𝑃 ↔ (𝑌‘0) = 𝑃))
32elrab 3559 . 2 (𝑌 ∈ {𝑤𝑍 ∣ (𝑤‘0) = 𝑃} ↔ (𝑌𝑍 ∧ (𝑌‘0) = 𝑃))
4 ibar 520 . . . . 5 ((𝑌‘0) = 𝑃 → ((𝜑𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑𝜓))))
5 3anass 1109 . . . . . 6 (((𝑌‘0) = 𝑃𝜑𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑𝜓)))
6 3ancoma 1112 . . . . . 6 (((𝑌‘0) = 𝑃𝜑𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓))
75, 6bitr3i 268 . . . . 5 (((𝑌‘0) = 𝑃 ∧ (𝜑𝜓)) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓))
84, 7syl6bb 278 . . . 4 ((𝑌‘0) = 𝑃 → ((𝜑𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)))
98ad2antlr 709 . . 3 (((𝑌𝑍 ∧ (𝑌‘0) = 𝑃) ∧ 𝑤𝑋) → ((𝜑𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)))
109rabbidva 3378 . 2 ((𝑌𝑍 ∧ (𝑌‘0) = 𝑃) → {𝑤𝑋 ∣ (𝜑𝜓)} = {𝑤𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)})
113, 10sylbi 208 1 (𝑌 ∈ {𝑤𝑍 ∣ (𝑤‘0) = 𝑃} → {𝑤𝑋 ∣ (𝜑𝜓)} = {𝑤𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)})
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 197  wa 384  w3a 1100   = wceq 1637  wcel 2156  {crab 3100  cfv 6097  0cc0 10217
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2068  ax-7 2104  ax-9 2165  ax-10 2185  ax-11 2201  ax-12 2214  ax-13 2420  ax-ext 2784
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-3an 1102  df-tru 1641  df-ex 1860  df-nf 1864  df-sb 2061  df-clab 2793  df-cleq 2799  df-clel 2802  df-nfc 2937  df-ral 3101  df-rex 3102  df-rab 3105  df-v 3393  df-uni 4631  df-br 4845  df-iota 6060  df-fv 6105
This theorem is referenced by:  rusgrnumwwlks  27115
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