Metamath Proof Explorer


Theorem ordsssuc2

Description: An ordinal subset of an ordinal number belongs to its successor. (Contributed by NM, 1-Feb-2005) (Proof shortened by Andrew Salmon, 12-Aug-2011)

Ref Expression
Assertion ordsssuc2 ( ( Ord 𝐴𝐵 ∈ On ) → ( 𝐴𝐵𝐴 ∈ suc 𝐵 ) )

Proof

Step Hyp Ref Expression
1 elong ( 𝐴 ∈ V → ( 𝐴 ∈ On ↔ Ord 𝐴 ) )
2 1 biimprd ( 𝐴 ∈ V → ( Ord 𝐴𝐴 ∈ On ) )
3 2 anim1d ( 𝐴 ∈ V → ( ( Ord 𝐴𝐵 ∈ On ) → ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ) )
4 onsssuc ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) → ( 𝐴𝐵𝐴 ∈ suc 𝐵 ) )
5 3 4 syl6 ( 𝐴 ∈ V → ( ( Ord 𝐴𝐵 ∈ On ) → ( 𝐴𝐵𝐴 ∈ suc 𝐵 ) ) )
6 annim ( ( 𝐵 ∈ On ∧ ¬ 𝐴 ∈ V ) ↔ ¬ ( 𝐵 ∈ On → 𝐴 ∈ V ) )
7 ssexg ( ( 𝐴𝐵𝐵 ∈ On ) → 𝐴 ∈ V )
8 7 ex ( 𝐴𝐵 → ( 𝐵 ∈ On → 𝐴 ∈ V ) )
9 elex ( 𝐴 ∈ suc 𝐵𝐴 ∈ V )
10 9 a1d ( 𝐴 ∈ suc 𝐵 → ( 𝐵 ∈ On → 𝐴 ∈ V ) )
11 8 10 pm5.21ni ( ¬ ( 𝐵 ∈ On → 𝐴 ∈ V ) → ( 𝐴𝐵𝐴 ∈ suc 𝐵 ) )
12 6 11 sylbi ( ( 𝐵 ∈ On ∧ ¬ 𝐴 ∈ V ) → ( 𝐴𝐵𝐴 ∈ suc 𝐵 ) )
13 12 expcom ( ¬ 𝐴 ∈ V → ( 𝐵 ∈ On → ( 𝐴𝐵𝐴 ∈ suc 𝐵 ) ) )
14 13 adantld ( ¬ 𝐴 ∈ V → ( ( Ord 𝐴𝐵 ∈ On ) → ( 𝐴𝐵𝐴 ∈ suc 𝐵 ) ) )
15 5 14 pm2.61i ( ( Ord 𝐴𝐵 ∈ On ) → ( 𝐴𝐵𝐴 ∈ suc 𝐵 ) )