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Theorem rusgrnumwwlkslem 29999
Description: Lemma for rusgrnumwwlks 30004. (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 6906 . . . 4 (𝑤 = 𝑌 → (𝑤‘0) = (𝑌‘0))
21eqeq1d 2737 . . 3 (𝑤 = 𝑌 → ((𝑤‘0) = 𝑃 ↔ (𝑌‘0) = 𝑃))
32elrab 3695 . 2 (𝑌 ∈ {𝑤𝑍 ∣ (𝑤‘0) = 𝑃} ↔ (𝑌𝑍 ∧ (𝑌‘0) = 𝑃))
4 ibar 528 . . . . 5 ((𝑌‘0) = 𝑃 → ((𝜑𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑𝜓))))
5 3anass 1094 . . . . . 6 (((𝑌‘0) = 𝑃𝜑𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑𝜓)))
6 3ancoma 1097 . . . . . 6 (((𝑌‘0) = 𝑃𝜑𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓))
75, 6bitr3i 277 . . . . 5 (((𝑌‘0) = 𝑃 ∧ (𝜑𝜓)) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓))
84, 7bitrdi 287 . . . 4 ((𝑌‘0) = 𝑃 → ((𝜑𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)))
98ad2antlr 727 . . 3 (((𝑌𝑍 ∧ (𝑌‘0) = 𝑃) ∧ 𝑤𝑋) → ((𝜑𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)))
109rabbidva 3440 . 2 ((𝑌𝑍 ∧ (𝑌‘0) = 𝑃) → {𝑤𝑋 ∣ (𝜑𝜓)} = {𝑤𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)})
113, 10sylbi 217 1 (𝑌 ∈ {𝑤𝑍 ∣ (𝑤‘0) = 𝑃} → {𝑤𝑋 ∣ (𝜑𝜓)} = {𝑤𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃𝜓)})
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395  w3a 1086   = wceq 1537  wcel 2106  {crab 3433  cfv 6563  0cc0 11153
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706
This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rab 3434  df-v 3480  df-ss 3980  df-uni 4913  df-br 5149  df-iota 6516  df-fv 6571
This theorem is referenced by:  rusgrnumwwlks  30004
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