Theorem rusgrnumwwlkslem 30002

Description: Lemma for rusgrnumwwlks 30007. (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 6919	. . . 4 ⊢ (𝑤 = 𝑌 → (𝑤‘0) = (𝑌‘0))
2	1	eqeq1d 2742	. . 3 ⊢ (𝑤 = 𝑌 → ((𝑤‘0) = 𝑃 ↔ (𝑌‘0) = 𝑃))
3	2	elrab 3708	. 2 ⊢ (𝑌 ∈ {𝑤 ∈ 𝑍 ∣ (𝑤‘0) = 𝑃} ↔ (𝑌 ∈ 𝑍 ∧ (𝑌‘0) = 𝑃))
4		ibar 528	. . . . 5 ⊢ ((𝑌‘0) = 𝑃 → ((𝜑 ∧ 𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑 ∧ 𝜓))))
5		3anass 1095	. . . . . 6 ⊢ (((𝑌‘0) = 𝑃 ∧ 𝜑 ∧ 𝜓) ↔ ((𝑌‘0) = 𝑃 ∧ (𝜑 ∧ 𝜓)))
6		3ancoma 1098	. . . . . 6 ⊢ (((𝑌‘0) = 𝑃 ∧ 𝜑 ∧ 𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓))
7	5, 6	bitr3i 277	. . . . 5 ⊢ (((𝑌‘0) = 𝑃 ∧ (𝜑 ∧ 𝜓)) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓))
8	4, 7	bitrdi 287	. . . 4 ⊢ ((𝑌‘0) = 𝑃 → ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)))
9	8	ad2antlr 726	. . 3 ⊢ (((𝑌 ∈ 𝑍 ∧ (𝑌‘0) = 𝑃) ∧ 𝑤 ∈ 𝑋) → ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)))
10	9	rabbidva 3450	. 2 ⊢ ((𝑌 ∈ 𝑍 ∧ (𝑌‘0) = 𝑃) → {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ 𝜓)} = {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)})
11	3, 10	sylbi 217	1 ⊢ (𝑌 ∈ {𝑤 ∈ 𝑍 ∣ (𝑤‘0) = 𝑃} → {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ 𝜓)} = {𝑤 ∈ 𝑋 ∣ (𝜑 ∧ (𝑌‘0) = 𝑃 ∧ 𝜓)})

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   ∧ w3a 1087   = wceq 1537   ∈ wcel 2108  {crab 3443  ‘cfv 6573  0cc0 11184

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1089  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-rab 3444  df-v 3490  df-ss 3993  df-uni 4932  df-br 5167  df-iota 6525  df-fv 6581

This theorem is referenced by:  rusgrnumwwlks  30007