Theorem phplem2 3404

Description: Lemma for Pigeonhole Principle. If we join a natural number to itself minus an element, we end up with its successor minus the same element.

Assertion

Ref	Expression
phplem2	⊢ ((A ∈ ω ∧ B ∈ A) → ({A} ∪ (A ∖ {B})) = (suc A ∖ {B}))

Proof of Theorem phplem2

Step	Hyp	Ref	Expression
1		nordeq 2218	. . . 4 ⊢ ((Ord A ∧ B ∈ A) → ¬ A = B)
2		disjsn2 1837	. . . 4 ⊢ (¬ A = B → ({A} ∩ {B}) = ∅)
3	1, 2	syl 12	. . 3 ⊢ ((Ord A ∧ B ∈ A) → ({A} ∩ {B}) = ∅)
4		nnord 2381	. . 3 ⊢ (A ∈ ω → Ord A)
5	3, 4	sylan 343	. 2 ⊢ ((A ∈ ω ∧ B ∈ A) → ({A} ∩ {B}) = ∅)
6		undif4 1744	. . 3 ⊢ (({A} ∩ {B}) = ∅ → ({A} ∪ (A ∖ {B})) = (({A} ∪ A) ∖ {B}))
7		df-suc 2205	. . . . 5 ⊢ suc A = (A ∪ {A})
8		uncom 1604	. . . . 5 ⊢ (A ∪ {A}) = ({A} ∪ A)
9	7, 8	eqtr 1119	. . . 4 ⊢ suc A = ({A} ∪ A)
10	9	difeq1i 1584	. . 3 ⊢ (suc A ∖ {B}) = (({A} ∪ A) ∖ {B})
11	6, 10	syl6eqr 1142	. 2 ⊢ (({A} ∩ {B}) = ∅ → ({A} ∪ (A ∖ {B})) = (suc A ∖ {B}))
12	5, 11	syl 12	1 ⊢ ((A ∈ ω ∧ B ∈ A) → ({A} ∪ (A ∖ {B})) = (suc A ∖ {B}))

Syntax hints:  ¬ wn 1   → wi 2   ∧ wa 196   = wceq 1091   ∈ wcel 1092   ∖ cdif 1484   ∪ cun 1485   ∩ cin 1486  ∅c0 1707  {csn 1808  Ord word 2198  suc csuc 2201  ωcom 2372

This theorem is referenced by:  phplem3 3405

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-13 804  ax-14 805  ax-16 922  ax-17 925  ax-ext 1074  ax-rep 1075  ax-pow 1077

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-3an 583  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811  df-pr 1812  df-op 1815  df-uni 1920  df-tr 2042  df-br 2063  df-opab 2098  df-eprel 2122  df-po 2128  df-so 2138  df-fr 2169  df-we 2186  df-ord 2202  df-on 2203  df-suc 2205  df-om 2373

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