Metamath Proof Explorer


Theorem oiiso

Description: The order isomorphism of the well-order R on A is an isomorphism. (Contributed by Mario Carneiro, 23-May-2015)

Ref Expression
Hypothesis oicl.1
|- F = OrdIso ( R , A )
Assertion oiiso
|- ( ( A e. V /\ R We A ) -> F Isom _E , R ( dom F , A ) )

Proof

Step Hyp Ref Expression
1 oicl.1
 |-  F = OrdIso ( R , A )
2 exse
 |-  ( A e. V -> R Se A )
3 1 ordtype
 |-  ( ( R We A /\ R Se A ) -> F Isom _E , R ( dom F , A ) )
4 3 ancoms
 |-  ( ( R Se A /\ R We A ) -> F Isom _E , R ( dom F , A ) )
5 2 4 sylan
 |-  ( ( A e. V /\ R We A ) -> F Isom _E , R ( dom F , A ) )