Metamath Proof Explorer


Theorem limsssuc

Description: A class includes a limit ordinal iff the successor of the class includes it. (Contributed by NM, 30-Oct-2003)

Ref Expression
Assertion limsssuc ( Lim 𝐴 → ( 𝐴 ⊆ 𝐵 ↔ 𝐴 ⊆ suc 𝐵 ) )

Proof

Step Hyp Ref Expression
1 sssucid ⊢ 𝐵 ⊆ suc 𝐵
2 sstr2 ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝐵 ⊆ suc 𝐵 → 𝐴 ⊆ suc 𝐵 ) )
3 1 2 mpi ⊢ ( 𝐴 ⊆ 𝐵 → 𝐴 ⊆ suc 𝐵 )
4 eleq1 ⊢ ( 𝑥 = 𝐵 → ( 𝑥 ∈ 𝐴 ↔ 𝐵 ∈ 𝐴 ) )
5 4 biimpcd ⊢ ( 𝑥 ∈ 𝐴 → ( 𝑥 = 𝐵 → 𝐵 ∈ 𝐴 ) )
6 limsuc ⊢ ( Lim 𝐴 → ( 𝐵 ∈ 𝐴 ↔ suc 𝐵 ∈ 𝐴 ) )
7 6 biimpa ⊢ ( ( Lim 𝐴 ∧ 𝐵 ∈ 𝐴 ) → suc 𝐵 ∈ 𝐴 )
8 limord ⊢ ( Lim 𝐴 → Ord 𝐴 )
9 ordelord ⊢ ( ( Ord 𝐴 ∧ 𝐵 ∈ 𝐴 ) → Ord 𝐵 )
10 8 9 sylan ⊢ ( ( Lim 𝐴 ∧ 𝐵 ∈ 𝐴 ) → Ord 𝐵 )
11 ordsuc ⊢ ( Ord 𝐵 ↔ Ord suc 𝐵 )
12 10 11 sylib ⊢ ( ( Lim 𝐴 ∧ 𝐵 ∈ 𝐴 ) → Ord suc 𝐵 )
13 ordtri1 ⊢ ( ( Ord 𝐴 ∧ Ord suc 𝐵 ) → ( 𝐴 ⊆ suc 𝐵 ↔ ¬ suc 𝐵 ∈ 𝐴 ) )
14 8 12 13 syl2an2r ⊢ ( ( Lim 𝐴 ∧ 𝐵 ∈ 𝐴 ) → ( 𝐴 ⊆ suc 𝐵 ↔ ¬ suc 𝐵 ∈ 𝐴 ) )
15 14 con2bid ⊢ ( ( Lim 𝐴 ∧ 𝐵 ∈ 𝐴 ) → ( suc 𝐵 ∈ 𝐴 ↔ ¬ 𝐴 ⊆ suc 𝐵 ) )
16 7 15 mpbid ⊢ ( ( Lim 𝐴 ∧ 𝐵 ∈ 𝐴 ) → ¬ 𝐴 ⊆ suc 𝐵 )
17 16 ex ⊢ ( Lim 𝐴 → ( 𝐵 ∈ 𝐴 → ¬ 𝐴 ⊆ suc 𝐵 ) )
18 5 17 sylan9r ⊢ ( ( Lim 𝐴 ∧ 𝑥 ∈ 𝐴 ) → ( 𝑥 = 𝐵 → ¬ 𝐴 ⊆ suc 𝐵 ) )
19 18 con2d ⊢ ( ( Lim 𝐴 ∧ 𝑥 ∈ 𝐴 ) → ( 𝐴 ⊆ suc 𝐵 → ¬ 𝑥 = 𝐵 ) )
20 19 ex ⊢ ( Lim 𝐴 → ( 𝑥 ∈ 𝐴 → ( 𝐴 ⊆ suc 𝐵 → ¬ 𝑥 = 𝐵 ) ) )
21 20 com23 ⊢ ( Lim 𝐴 → ( 𝐴 ⊆ suc 𝐵 → ( 𝑥 ∈ 𝐴 → ¬ 𝑥 = 𝐵 ) ) )
22 21 imp31 ⊢ ( ( ( Lim 𝐴 ∧ 𝐴 ⊆ suc 𝐵 ) ∧ 𝑥 ∈ 𝐴 ) → ¬ 𝑥 = 𝐵 )
23 ssel2 ⊢ ( ( 𝐴 ⊆ suc 𝐵 ∧ 𝑥 ∈ 𝐴 ) → 𝑥 ∈ suc 𝐵 )
24 vex ⊢ 𝑥 ∈ V
25 24 elsuc ⊢ ( 𝑥 ∈ suc 𝐵 ↔ ( 𝑥 ∈ 𝐵 ∨ 𝑥 = 𝐵 ) )
26 23 25 sylib ⊢ ( ( 𝐴 ⊆ suc 𝐵 ∧ 𝑥 ∈ 𝐴 ) → ( 𝑥 ∈ 𝐵 ∨ 𝑥 = 𝐵 ) )
27 26 ord ⊢ ( ( 𝐴 ⊆ suc 𝐵 ∧ 𝑥 ∈ 𝐴 ) → ( ¬ 𝑥 ∈ 𝐵 → 𝑥 = 𝐵 ) )
28 27 con1d ⊢ ( ( 𝐴 ⊆ suc 𝐵 ∧ 𝑥 ∈ 𝐴 ) → ( ¬ 𝑥 = 𝐵 → 𝑥 ∈ 𝐵 ) )
29 28 adantll ⊢ ( ( ( Lim 𝐴 ∧ 𝐴 ⊆ suc 𝐵 ) ∧ 𝑥 ∈ 𝐴 ) → ( ¬ 𝑥 = 𝐵 → 𝑥 ∈ 𝐵 ) )
30 22 29 mpd ⊢ ( ( ( Lim 𝐴 ∧ 𝐴 ⊆ suc 𝐵 ) ∧ 𝑥 ∈ 𝐴 ) → 𝑥 ∈ 𝐵 )
31 30 ex ⊢ ( ( Lim 𝐴 ∧ 𝐴 ⊆ suc 𝐵 ) → ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
32 31 ssrdv ⊢ ( ( Lim 𝐴 ∧ 𝐴 ⊆ suc 𝐵 ) → 𝐴 ⊆ 𝐵 )
33 32 ex ⊢ ( Lim 𝐴 → ( 𝐴 ⊆ suc 𝐵 → 𝐴 ⊆ 𝐵 ) )
34 3 33 impbid2 ⊢ ( Lim 𝐴 → ( 𝐴 ⊆ 𝐵 ↔ 𝐴 ⊆ suc 𝐵 ) )