Theorem 1wlkdlem2 27558

Description: Lemma 2 for 1wlkd 27561. (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 4428	. . . . 5 ⊢ (𝑋 ∈ 𝑉 → 𝑋 ∈ {𝑋})
3	1, 2	syl 17	. . . 4 ⊢ (𝜑 → 𝑋 ∈ {𝑋})
4	3	adantr 474	. . 3 ⊢ ((𝜑 ∧ 𝑋 = 𝑌) → 𝑋 ∈ {𝑋})
5		1wlkd.l	. . 3 ⊢ ((𝜑 ∧ 𝑋 = 𝑌) → (𝐼‘𝐽) = {𝑋})
6	4, 5	eleqtrrd 2862	. 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 474	. . . . 5 ⊢ ((𝜑 ∧ 𝑋 ≠ 𝑌) → 𝑌 ∈ 𝑉)
10		prssg 4583	. . . . 5 ⊢ ((𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → ((𝑋 ∈ (𝐼‘𝐽) ∧ 𝑌 ∈ (𝐼‘𝐽)) ↔ {𝑋, 𝑌} ⊆ (𝐼‘𝐽)))
11	1, 9, 10	syl2an2r 675	. . . 4 ⊢ ((𝜑 ∧ 𝑋 ≠ 𝑌) → ((𝑋 ∈ (𝐼‘𝐽) ∧ 𝑌 ∈ (𝐼‘𝐽)) ↔ {𝑋, 𝑌} ⊆ (𝐼‘𝐽)))
12	7, 11	mpbird 249	. . 3 ⊢ ((𝜑 ∧ 𝑋 ≠ 𝑌) → (𝑋 ∈ (𝐼‘𝐽) ∧ 𝑌 ∈ (𝐼‘𝐽)))
13	12	simpld 490	. 2 ⊢ ((𝜑 ∧ 𝑋 ≠ 𝑌) → 𝑋 ∈ (𝐼‘𝐽))
14	6, 13	pm2.61dane 3057	1 ⊢ (𝜑 → 𝑋 ∈ (𝐼‘𝐽))

Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 386   = wceq 1601   ∈ wcel 2107   ≠ wne 2969   ⊆ wss 3792  {csn 4398  {cpr 4400  ‘cfv 6137  ⟨“cs1 13691  ⟨“cs2 13998

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1839  ax-4 1853  ax-5 1953  ax-6 2021  ax-7 2055  ax-9 2116  ax-10 2135  ax-11 2150  ax-12 2163  ax-ext 2754

This theorem depends on definitions:  df-bi 199  df-an 387  df-or 837  df-3an 1073  df-tru 1605  df-ex 1824  df-nf 1828  df-sb 2012  df-clab 2764  df-cleq 2770  df-clel 2774  df-nfc 2921  df-ne 2970  df-v 3400  df-un 3797  df-in 3799  df-ss 3806  df-sn 4399  df-pr 4401

This theorem is referenced by:  1wlkdlem3  27559  1wlkdlem4  27560