Metamath Proof Explorer


Theorem oldss

Description: If A is less than or equal to ordinal B , then the old set of A is included in the made set of B . (Contributed by Scott Fenton, 9-Oct-2024)

Ref Expression
Assertion oldss ( ( 𝐵 ∈ On ∧ 𝐴𝐵 ) → ( O ‘ 𝐴 ) ⊆ ( O ‘ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 imass2 ( 𝐴𝐵 → ( M “ 𝐴 ) ⊆ ( M “ 𝐵 ) )
2 1 unissd ( 𝐴𝐵 ( M “ 𝐴 ) ⊆ ( M “ 𝐵 ) )
3 2 adantl ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ 𝐴𝐵 ) → ( M “ 𝐴 ) ⊆ ( M “ 𝐵 ) )
4 oldval ( 𝐴 ∈ On → ( O ‘ 𝐴 ) = ( M “ 𝐴 ) )
5 4 adantr ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) → ( O ‘ 𝐴 ) = ( M “ 𝐴 ) )
6 5 adantr ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ 𝐴𝐵 ) → ( O ‘ 𝐴 ) = ( M “ 𝐴 ) )
7 oldval ( 𝐵 ∈ On → ( O ‘ 𝐵 ) = ( M “ 𝐵 ) )
8 7 adantl ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) → ( O ‘ 𝐵 ) = ( M “ 𝐵 ) )
9 8 adantr ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ 𝐴𝐵 ) → ( O ‘ 𝐵 ) = ( M “ 𝐵 ) )
10 3 6 9 3sstr4d ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ 𝐴𝐵 ) → ( O ‘ 𝐴 ) ⊆ ( O ‘ 𝐵 ) )
11 10 expl ( 𝐴 ∈ On → ( ( 𝐵 ∈ On ∧ 𝐴𝐵 ) → ( O ‘ 𝐴 ) ⊆ ( O ‘ 𝐵 ) ) )
12 oldf O : On ⟶ 𝒫 No
13 12 fdmi dom O = On
14 13 eleq2i ( 𝐴 ∈ dom O ↔ 𝐴 ∈ On )
15 ndmfv ( ¬ 𝐴 ∈ dom O → ( O ‘ 𝐴 ) = ∅ )
16 14 15 sylnbir ( ¬ 𝐴 ∈ On → ( O ‘ 𝐴 ) = ∅ )
17 0ss ∅ ⊆ ( O ‘ 𝐵 )
18 16 17 eqsstrdi ( ¬ 𝐴 ∈ On → ( O ‘ 𝐴 ) ⊆ ( O ‘ 𝐵 ) )
19 18 a1d ( ¬ 𝐴 ∈ On → ( ( 𝐵 ∈ On ∧ 𝐴𝐵 ) → ( O ‘ 𝐴 ) ⊆ ( O ‘ 𝐵 ) ) )
20 11 19 pm2.61i ( ( 𝐵 ∈ On ∧ 𝐴𝐵 ) → ( O ‘ 𝐴 ) ⊆ ( O ‘ 𝐵 ) )