Theorem kur14lem1 36138

Description: Lemma for kur14 36148. (Contributed by Mario Carneiro, 17-Feb-2015.)

Hypotheses

Ref	Expression
kur14lem1.a	⊢ 𝐴 ⊆ 𝑋
kur14lem1.c	⊢ (𝑋 ∖ 𝐴) ∈ 𝑇
kur14lem1.k	⊢ (𝐾‘𝐴) ∈ 𝑇

Assertion

Ref	Expression
kur14lem1	⊢ (𝑁 = 𝐴 → (𝑁 ⊆ 𝑋 ∧ {(𝑋 ∖ 𝑁), (𝐾‘𝑁)} ⊆ 𝑇))

Proof of Theorem kur14lem1

Step	Hyp	Ref	Expression
1		kur14lem1.a	. . 3 ⊢ 𝐴 ⊆ 𝑋
2		sseq1 3938	. . 3 ⊢ (𝑁 = 𝐴 → (𝑁 ⊆ 𝑋 ↔ 𝐴 ⊆ 𝑋))
3	1, 2	mpbiri 261	. 2 ⊢ (𝑁 = 𝐴 → 𝑁 ⊆ 𝑋)
4		difeq2 4060	. . . 4 ⊢ (𝑁 = 𝐴 → (𝑋 ∖ 𝑁) = (𝑋 ∖ 𝐴))
5		fveq2 6908	. . . 4 ⊢ (𝑁 = 𝐴 → (𝐾‘𝑁) = (𝐾‘𝐴))
6	4, 5	preq12d 4702	. . 3 ⊢ (𝑁 = 𝐴 → {(𝑋 ∖ 𝑁), (𝐾‘𝑁)} = {(𝑋 ∖ 𝐴), (𝐾‘𝐴)})
7		kur14lem1.c	. . . 4 ⊢ (𝑋 ∖ 𝐴) ∈ 𝑇
8		kur14lem1.k	. . . 4 ⊢ (𝐾‘𝐴) ∈ 𝑇
9		prssi 4780	. . . 4 ⊢ (((𝑋 ∖ 𝐴) ∈ 𝑇 ∧ (𝐾‘𝐴) ∈ 𝑇) → {(𝑋 ∖ 𝐴), (𝐾‘𝐴)} ⊆ 𝑇)
10	7, 8, 9	mp2an 705	. . 3 ⊢ {(𝑋 ∖ 𝐴), (𝐾‘𝐴)} ⊆ 𝑇
11	6, 10	eqsstrdi 3957	. 2 ⊢ (𝑁 = 𝐴 → {(𝑋 ∖ 𝑁), (𝐾‘𝑁)} ⊆ 𝑇)
12	3, 11	jca 521	1 ⊢ (𝑁 = 𝐴 → (𝑁 ⊆ 𝑋 ∧ {(𝑋 ∖ 𝑁), (𝐾‘𝑁)} ⊆ 𝑇))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2147   ⊆ wss 3900   ∖ cdif 4048  {cpr 4590  ‘cfv 6558

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-uni 4871  df-br 5107  df-iota 6512  df-fv 6566

This theorem is used by:  kur14lem7  36144