Theorem rusgrnumwwlkslem 28025

Description: Lemma for rusgrnumwwlks 28030. (Contributed by Alexander van der Vekens, 23-Aug-2018.)

Assertion

Ref	Expression
rusgrnumwwlkslem	⊢ (𝑌 ∈ {𝑤 ∈ 𝑍 ∣ (𝑤‘0) = 𝑃} → {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ 𝜓)} = {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)})

Distinct variable groups:   𝑤,𝑃   𝑤,𝑌   𝑤,𝑍

Allowed substitution hints:   𝜑(𝑤)   𝜓(𝑤)   𝑋(𝑤)

Proof of Theorem rusgrnumwwlkslem

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)})

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