Metamath Proof Explorer


Theorem oiexg

Description: The order isomorphism on a set is a set. (Contributed by Mario Carneiro, 25-Jun-2015)

Ref Expression
Hypothesis oicl.1 𝐹 = OrdIso ( 𝑅 , 𝐴 )
Assertion oiexg ( 𝐴𝑉𝐹 ∈ V )

Proof

Step Hyp Ref Expression
1 oicl.1 𝐹 = OrdIso ( 𝑅 , 𝐴 )
2 1 ordtype ( ( 𝑅 We 𝐴𝑅 Se 𝐴 ) → 𝐹 Isom E , 𝑅 ( dom 𝐹 , 𝐴 ) )
3 isof1o ( 𝐹 Isom E , 𝑅 ( dom 𝐹 , 𝐴 ) → 𝐹 : dom 𝐹1-1-onto𝐴 )
4 f1of1 ( 𝐹 : dom 𝐹1-1-onto𝐴𝐹 : dom 𝐹1-1𝐴 )
5 2 3 4 3syl ( ( 𝑅 We 𝐴𝑅 Se 𝐴 ) → 𝐹 : dom 𝐹1-1𝐴 )
6 f1f ( 𝐹 : dom 𝐹1-1𝐴𝐹 : dom 𝐹𝐴 )
7 f1dmex ( ( 𝐹 : dom 𝐹1-1𝐴𝐴𝑉 ) → dom 𝐹 ∈ V )
8 fex ( ( 𝐹 : dom 𝐹𝐴 ∧ dom 𝐹 ∈ V ) → 𝐹 ∈ V )
9 6 7 8 syl2an2r ( ( 𝐹 : dom 𝐹1-1𝐴𝐴𝑉 ) → 𝐹 ∈ V )
10 9 expcom ( 𝐴𝑉 → ( 𝐹 : dom 𝐹1-1𝐴𝐹 ∈ V ) )
11 5 10 syl5 ( 𝐴𝑉 → ( ( 𝑅 We 𝐴𝑅 Se 𝐴 ) → 𝐹 ∈ V ) )
12 1 oi0 ( ¬ ( 𝑅 We 𝐴𝑅 Se 𝐴 ) → 𝐹 = ∅ )
13 0ex ∅ ∈ V
14 12 13 eqeltrdi ( ¬ ( 𝑅 We 𝐴𝑅 Se 𝐴 ) → 𝐹 ∈ V )
15 11 14 pm2.61d1 ( 𝐴𝑉𝐹 ∈ V )