Theorem ordequn 2331

Description: The maximum (i.e. union) of two ordinals is either one or the other. Similar to Exercise 14 of [TakeutiZaring] p. 40.

Assertion

Ref	Expression
ordequn	⊢ ((Ord B ∧ Ord C) → (A = (B ∪ C) → (A = B ∨ A = C)))

Proof of Theorem ordequn

Step	Hyp	Ref	Expression
1		ordtri2or2 2329	. 2 ⊢ ((Ord B ∧ Ord C) → (B ⊆ C ∨ C ⊆ B))
2		ssequn1 1628	. . . . 5 ⊢ (B ⊆ C ↔ (B ∪ C) = C)
3		cleq2 1110	. . . . 5 ⊢ ((B ∪ C) = C → (A = (B ∪ C) ↔ A = C))
4	2, 3	sylbi 174	. . . 4 ⊢ (B ⊆ C → (A = (B ∪ C) ↔ A = C))
5		olc 224	. . . 4 ⊢ (A = C → (A = B ∨ A = C))
6	4, 5	syl6bi 187	. . 3 ⊢ (B ⊆ C → (A = (B ∪ C) → (A = B ∨ A = C)))
7		ssequn2 1631	. . . . 5 ⊢ (C ⊆ B ↔ (B ∪ C) = B)
8		cleq2 1110	. . . . 5 ⊢ ((B ∪ C) = B → (A = (B ∪ C) ↔ A = B))
9	7, 8	sylbi 174	. . . 4 ⊢ (C ⊆ B → (A = (B ∪ C) ↔ A = B))
10		orc 225	. . . 4 ⊢ (A = B → (A = B ∨ A = C))
11	9, 10	syl6bi 187	. . 3 ⊢ (C ⊆ B → (A = (B ∪ C) → (A = B ∨ A = C)))
12	6, 11	jaoi 275	. 2 ⊢ ((B ⊆ C ∨ C ⊆ B) → (A = (B ∪ C) → (A = B ∨ A = C)))
13	1, 12	syl 12	1 ⊢ ((Ord B ∧ Ord C) → (A = (B ∪ C) → (A = B ∨ A = C)))

Syntax hints:   → wi 2   ↔ wb 127   ∨ wo 195   ∧ wa 196   = wceq 1091   ∪ cun 1485   ⊆ wss 1487  Ord word 2198

This theorem is referenced by:  ordun 2332

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-3or 582  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

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