Theorem 1wlkdlem2 30398

Description: Lemma 2 for 1wlkd 30401. (Contributed by AV, 22-Jan-2021.)

Hypotheses

Ref	Expression
1wlkd.p	⊢ 𝑃 = ⟨“𝑋𝑌”⟩
1wlkd.f	⊢ 𝐹 = ⟨“𝐽”⟩
1wlkd.x	⊢ (𝜑 → 𝑋 ∈ 𝑉)
1wlkd.y	⊢ (𝜑 → 𝑌 ∈ 𝑉)
1wlkd.l	⊢ ((𝜑 ∧ 𝑋 = 𝑌) → (𝐼‘𝐽) = {𝑋})
1wlkd.j	⊢ ((𝜑 ∧ 𝑋 ≠ 𝑌) → {𝑋, 𝑌} ⊆ (𝐼‘𝐽))

Assertion

Ref	Expression
1wlkdlem2	⊢ (𝜑 → 𝑋 ∈ (𝐼‘𝐽))

Proof of Theorem 1wlkdlem2

Step	Hyp	Ref	Expression
1		1wlkd.x	. . . . 5 ⊢ (𝜑 → 𝑋 ∈ 𝑉)
2		snidg 4622	. . . . 5 ⊢ (𝑋 ∈ 𝑉 → 𝑋 ∈ {𝑋})
3	1, 2	syl 18	. . . 4 ⊢ (𝜑 → 𝑋 ∈ {𝑋})
4	3	adantr 485	. . 3 ⊢ ((𝜑 ∧ 𝑋 = 𝑌) → 𝑋 ∈ {𝑋})
5		1wlkd.l	. . 3 ⊢ ((𝜑 ∧ 𝑋 = 𝑌) → (𝐼‘𝐽) = {𝑋})
6	4, 5	eleqtrrd 2868	. 2 ⊢ ((𝜑 ∧ 𝑋 = 𝑌) → 𝑋 ∈ (𝐼‘𝐽))
7		1wlkd.j	. . . 4 ⊢ ((𝜑 ∧ 𝑋 ≠ 𝑌) → {𝑋, 𝑌} ⊆ (𝐼‘𝐽))
8		1wlkd.y	. . . . . 6 ⊢ (𝜑 → 𝑌 ∈ 𝑉)
9	8	adantr 485	. . . . 5 ⊢ ((𝜑 ∧ 𝑋 ≠ 𝑌) → 𝑌 ∈ 𝑉)
10		prssg 4780	. . . . 5 ⊢ ((𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → ((𝑋 ∈ (𝐼‘𝐽) ∧ 𝑌 ∈ (𝐼‘𝐽)) ↔ {𝑋, 𝑌} ⊆ (𝐼‘𝐽)))
11	1, 9, 10	syl2an2r 697	. . . 4 ⊢ ((𝜑 ∧ 𝑋 ≠ 𝑌) → ((𝑋 ∈ (𝐼‘𝐽) ∧ 𝑌 ∈ (𝐼‘𝐽)) ↔ {𝑋, 𝑌} ⊆ (𝐼‘𝐽)))
12	7, 11	mpbird 260	. . 3 ⊢ ((𝜑 ∧ 𝑋 ≠ 𝑌) → (𝑋 ∈ (𝐼‘𝐽) ∧ 𝑌 ∈ (𝐼‘𝐽)))
13	12	simpld 499	. 2 ⊢ ((𝜑 ∧ 𝑋 ≠ 𝑌) → 𝑋 ∈ (𝐼‘𝐽))
14	6, 13	pm2.61dane 3047	1 ⊢ (𝜑 → 𝑋 ∈ (𝐼‘𝐽))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1563   ∈ wcel 2145   ≠ wne 2960   ⊆ wss 3907  {csn 4585  {cpr 4587  ‘cfv 6525  ⟨“cs1 14623  ⟨“cs2 14868

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-ext 2737

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1566  df-ex 1803  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-v 3459  df-un 3912  df-ss 3924  df-sn 4586  df-pr 4588

This theorem is referenced by:  1wlkdlem3  30399  1wlkdlem4  30400