Theorem ordsucun 7825

Description: The successor of the maximum (i.e. union) of two ordinals is the maximum of their successors. (Contributed by NM, 28-Nov-2003.)

Assertion

Ref	Expression
ordsucun	⊢ ((Ord 𝐴 ∧ Ord 𝐵) → suc (𝐴 ∪ 𝐵) = (suc 𝐴 ∪ suc 𝐵))

Proof of Theorem ordsucun

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ordun 6468	. . . 4 ⊢ ((Ord 𝐴 ∧ Ord 𝐵) → Ord (𝐴 ∪ 𝐵))
2		ordsuc 7814	. . . . 5 ⊢ (Ord (𝐴 ∪ 𝐵) ↔ Ord suc (𝐴 ∪ 𝐵))
3		ordelon 6385	. . . . . 6 ⊢ ((Ord suc (𝐴 ∪ 𝐵) ∧ 𝑥 ∈ suc (𝐴 ∪ 𝐵)) → 𝑥 ∈ On)
4	3	ex 418	. . . . 5 ⊢ (Ord suc (𝐴 ∪ 𝐵) → (𝑥 ∈ suc (𝐴 ∪ 𝐵) → 𝑥 ∈ On))
5	2, 4	sylbi 220	. . . 4 ⊢ (Ord (𝐴 ∪ 𝐵) → (𝑥 ∈ suc (𝐴 ∪ 𝐵) → 𝑥 ∈ On))
6	1, 5	syl 18	. . 3 ⊢ ((Ord 𝐴 ∧ Ord 𝐵) → (𝑥 ∈ suc (𝐴 ∪ 𝐵) → 𝑥 ∈ On))
7		ordsuc 7814	. . . 4 ⊢ (Ord 𝐴 ↔ Ord suc 𝐴)
8		ordsuc 7814	. . . 4 ⊢ (Ord 𝐵 ↔ Ord suc 𝐵)
9		ordun 6468	. . . . 5 ⊢ ((Ord suc 𝐴 ∧ Ord suc 𝐵) → Ord (suc 𝐴 ∪ suc 𝐵))
10		ordelon 6385	. . . . . 6 ⊢ ((Ord (suc 𝐴 ∪ suc 𝐵) ∧ 𝑥 ∈ (suc 𝐴 ∪ suc 𝐵)) → 𝑥 ∈ On)
11	10	ex 418	. . . . 5 ⊢ (Ord (suc 𝐴 ∪ suc 𝐵) → (𝑥 ∈ (suc 𝐴 ∪ suc 𝐵) → 𝑥 ∈ On))
12	9, 11	syl 18	. . . 4 ⊢ ((Ord suc 𝐴 ∧ Ord suc 𝐵) → (𝑥 ∈ (suc 𝐴 ∪ suc 𝐵) → 𝑥 ∈ On))
13	7, 8, 12	syl2anb 610	. . 3 ⊢ ((Ord 𝐴 ∧ Ord 𝐵) → (𝑥 ∈ (suc 𝐴 ∪ suc 𝐵) → 𝑥 ∈ On))
14		ordssun 6466	. . . . . . 7 ⊢ ((Ord 𝐴 ∧ Ord 𝐵) → (𝑥 ⊆ (𝐴 ∪ 𝐵) ↔ (𝑥 ⊆ 𝐴 ∨ 𝑥 ⊆ 𝐵)))
15	14	adantl 487	. . . . . 6 ⊢ ((𝑥 ∈ On ∧ (Ord 𝐴 ∧ Ord 𝐵)) → (𝑥 ⊆ (𝐴 ∪ 𝐵) ↔ (𝑥 ⊆ 𝐴 ∨ 𝑥 ⊆ 𝐵)))
16		ordsssuc 6453	. . . . . . 7 ⊢ ((𝑥 ∈ On ∧ Ord (𝐴 ∪ 𝐵)) → (𝑥 ⊆ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ suc (𝐴 ∪ 𝐵)))
17	1, 16	sylan2 605	. . . . . 6 ⊢ ((𝑥 ∈ On ∧ (Ord 𝐴 ∧ Ord 𝐵)) → (𝑥 ⊆ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ suc (𝐴 ∪ 𝐵)))
18		ordsssuc 6453	. . . . . . . 8 ⊢ ((𝑥 ∈ On ∧ Ord 𝐴) → (𝑥 ⊆ 𝐴 ↔ 𝑥 ∈ suc 𝐴))
19	18	adantrr 730	. . . . . . 7 ⊢ ((𝑥 ∈ On ∧ (Ord 𝐴 ∧ Ord 𝐵)) → (𝑥 ⊆ 𝐴 ↔ 𝑥 ∈ suc 𝐴))
20		ordsssuc 6453	. . . . . . . 8 ⊢ ((𝑥 ∈ On ∧ Ord 𝐵) → (𝑥 ⊆ 𝐵 ↔ 𝑥 ∈ suc 𝐵))
21	20	adantrl 729	. . . . . . 7 ⊢ ((𝑥 ∈ On ∧ (Ord 𝐴 ∧ Ord 𝐵)) → (𝑥 ⊆ 𝐵 ↔ 𝑥 ∈ suc 𝐵))
22	19, 21	orbi12d 932	. . . . . 6 ⊢ ((𝑥 ∈ On ∧ (Ord 𝐴 ∧ Ord 𝐵)) → ((𝑥 ⊆ 𝐴 ∨ 𝑥 ⊆ 𝐵) ↔ (𝑥 ∈ suc 𝐴 ∨ 𝑥 ∈ suc 𝐵)))
23	15, 17, 22	3bitr3d 312	. . . . 5 ⊢ ((𝑥 ∈ On ∧ (Ord 𝐴 ∧ Ord 𝐵)) → (𝑥 ∈ suc (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ suc 𝐴 ∨ 𝑥 ∈ suc 𝐵)))
24		elun 4103	. . . . 5 ⊢ (𝑥 ∈ (suc 𝐴 ∪ suc 𝐵) ↔ (𝑥 ∈ suc 𝐴 ∨ 𝑥 ∈ suc 𝐵))
25	23, 24	bitr4di 292	. . . 4 ⊢ ((𝑥 ∈ On ∧ (Ord 𝐴 ∧ Ord 𝐵)) → (𝑥 ∈ suc (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ (suc 𝐴 ∪ suc 𝐵)))
26	25	expcom 419	. . 3 ⊢ ((Ord 𝐴 ∧ Ord 𝐵) → (𝑥 ∈ On → (𝑥 ∈ suc (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ (suc 𝐴 ∪ suc 𝐵))))
27	6, 13, 26	pm5.21ndd 382	. 2 ⊢ ((Ord 𝐴 ∧ Ord 𝐵) → (𝑥 ∈ suc (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ (suc 𝐴 ∪ suc 𝐵)))
28	27	eqrdv 2760	1 ⊢ ((Ord 𝐴 ∧ Ord 𝐵) → suc (𝐴 ∪ 𝐵) = (suc 𝐴 ∪ suc 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1570   ∈ wcel 2145   ∪ cun 3900   ⊆ wss 3902  Ord word 6360  Oncon0 6361  suc csuc 6363

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734  ax-sep 5255  ax-pr 5402

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3or 1104  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-pss 3922  df-nul 4283  df-if 4486  df-pw 4562  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4871  df-br 5108  df-opab 5172  df-tr 5217  df-eprel 5559  df-po 5567  df-so 5568  df-fr 5612  df-we 5614  df-ord 6364  df-on 6365  df-suc 6367

This theorem is used by:  rankprb  9837