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Theorem rusgrnumwwlkslem 26858
Description: Lemma for rusgrnumwwlks 26863. (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 6188 . . . 4 (𝑤 = 𝑌 → (𝑤‘0) = (𝑌‘0))
21eqeq1d 2623 . . 3 (𝑤 = 𝑌 → ((𝑤‘0) = 𝑃 ↔ (𝑌‘0) = 𝑃))
32elrab 3361 . 2 (𝑌 ∈ {𝑤𝑍 ∣ (𝑤‘0) = 𝑃} ↔ (𝑌𝑍 ∧ (𝑌‘0) = 𝑃))
4 ibar 525 . . . . 5 ((𝑌‘0) = 𝑃 → ((𝜑𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑𝜓))))
5 3anass 1041 . . . . . 6 (((𝑌‘0) = 𝑃𝜑𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑𝜓)))
6 3ancoma 1044 . . . . . 6 (((𝑌‘0) = 𝑃𝜑𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓))
75, 6bitr3i 266 . . . . 5 (((𝑌‘0) = 𝑃 ∧ (𝜑𝜓)) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓))
84, 7syl6bb 276 . . . 4 ((𝑌‘0) = 𝑃 → ((𝜑𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)))
98ad2antlr 763 . . 3 (((𝑌𝑍 ∧ (𝑌‘0) = 𝑃) ∧ 𝑤𝑋) → ((𝜑𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)))
109rabbidva 3186 . 2 ((𝑌𝑍 ∧ (𝑌‘0) = 𝑃) → {𝑤𝑋 ∣ (𝜑𝜓)} = {𝑤𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)})
113, 10sylbi 207 1 (𝑌 ∈ {𝑤𝑍 ∣ (𝑤‘0) = 𝑃} → {𝑤𝑋 ∣ (𝜑𝜓)} = {𝑤𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)})
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wa 384  w3a 1037   = wceq 1482  wcel 1989  {crab 2915  cfv 5886  0cc0 9933
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1721  ax-4 1736  ax-5 1838  ax-6 1887  ax-7 1934  ax-9 1998  ax-10 2018  ax-11 2033  ax-12 2046  ax-13 2245  ax-ext 2601
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1485  df-ex 1704  df-nf 1709  df-sb 1880  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2752  df-ral 2916  df-rex 2917  df-rab 2920  df-v 3200  df-uni 4435  df-br 4652  df-iota 5849  df-fv 5894
This theorem is referenced by:  rusgrnumwwlks  26863
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