Metamath Proof Explorer


Theorem ordeldifsucon

Description: Membership in the difference of ordinal and successor ordinal. (Contributed by RP, 16-Jan-2025)

Ref Expression
Assertion ordeldifsucon ( ( Ord 𝐴 ∧ 𝐵 ∈ On ) → ( 𝐶 ∈ ( 𝐴 ∖ suc 𝐵 ) ↔ ( 𝐶 ∈ 𝐴 ∧ 𝐵 ∈ 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 eldif ⊢ ( 𝐶 ∈ ( 𝐴 ∖ suc 𝐵 ) ↔ ( 𝐶 ∈ 𝐴 ∧ ¬ 𝐶 ∈ suc 𝐵 ) )
2 simplr ⊢ ( ( ( Ord 𝐴 ∧ 𝐵 ∈ On ) ∧ 𝐶 ∈ 𝐴 ) → 𝐵 ∈ On )
3 ordelord ⊢ ( ( Ord 𝐴 ∧ 𝐶 ∈ 𝐴 ) → Ord 𝐶 )
4 3 adantlr ⊢ ( ( ( Ord 𝐴 ∧ 𝐵 ∈ On ) ∧ 𝐶 ∈ 𝐴 ) → Ord 𝐶 )
5 ordelsuc ⊢ ( ( 𝐵 ∈ On ∧ Ord 𝐶 ) → ( 𝐵 ∈ 𝐶 ↔ suc 𝐵 ⊆ 𝐶 ) )
6 2 4 5 syl2anc ⊢ ( ( ( Ord 𝐴 ∧ 𝐵 ∈ On ) ∧ 𝐶 ∈ 𝐴 ) → ( 𝐵 ∈ 𝐶 ↔ suc 𝐵 ⊆ 𝐶 ) )
7 eloni ⊢ ( 𝐵 ∈ On → Ord 𝐵 )
8 ordsuci ⊢ ( Ord 𝐵 → Ord suc 𝐵 )
9 2 7 8 3syl ⊢ ( ( ( Ord 𝐴 ∧ 𝐵 ∈ On ) ∧ 𝐶 ∈ 𝐴 ) → Ord suc 𝐵 )
10 ordtri1 ⊢ ( ( Ord suc 𝐵 ∧ Ord 𝐶 ) → ( suc 𝐵 ⊆ 𝐶 ↔ ¬ 𝐶 ∈ suc 𝐵 ) )
11 9 4 10 syl2anc ⊢ ( ( ( Ord 𝐴 ∧ 𝐵 ∈ On ) ∧ 𝐶 ∈ 𝐴 ) → ( suc 𝐵 ⊆ 𝐶 ↔ ¬ 𝐶 ∈ suc 𝐵 ) )
12 6 11 bitr2d ⊢ ( ( ( Ord 𝐴 ∧ 𝐵 ∈ On ) ∧ 𝐶 ∈ 𝐴 ) → ( ¬ 𝐶 ∈ suc 𝐵 ↔ 𝐵 ∈ 𝐶 ) )
13 12 pm5.32da ⊢ ( ( Ord 𝐴 ∧ 𝐵 ∈ On ) → ( ( 𝐶 ∈ 𝐴 ∧ ¬ 𝐶 ∈ suc 𝐵 ) ↔ ( 𝐶 ∈ 𝐴 ∧ 𝐵 ∈ 𝐶 ) ) )
14 1 13 bitrid ⊢ ( ( Ord 𝐴 ∧ 𝐵 ∈ On ) → ( 𝐶 ∈ ( 𝐴 ∖ suc 𝐵 ) ↔ ( 𝐶 ∈ 𝐴 ∧ 𝐵 ∈ 𝐶 ) ) )