Metamath Proof Explorer


Theorem smoiso

Description: If F is an isomorphism from an ordinal A onto B , which is a subset of the ordinals, then F is a strictly monotonic function. Exercise 3 in TakeutiZaring p. 50. (Contributed by Andrew Salmon, 24-Nov-2011)

Ref Expression
Assertion smoiso F Isom E , E A B Ord A B On Smo F

Proof

Step Hyp Ref Expression
1 isof1o F Isom E , E A B F : A 1-1 onto B
2 f1of F : A 1-1 onto B F : A B
3 1 2 syl F Isom E , E A B F : A B
4 ffdm F : A B F : dom F B dom F A
5 4 simpld F : A B F : dom F B
6 fss F : dom F B B On F : dom F On
7 5 6 sylan F : A B B On F : dom F On
8 7 3adant2 F : A B Ord A B On F : dom F On
9 3 8 syl3an1 F Isom E , E A B Ord A B On F : dom F On
10 fdm F : A B dom F = A
11 10 eqcomd F : A B A = dom F
12 ordeq A = dom F Ord A Ord dom F
13 1 2 11 12 4syl F Isom E , E A B Ord A Ord dom F
14 13 biimpa F Isom E , E A B Ord A Ord dom F
15 14 3adant3 F Isom E , E A B Ord A B On Ord dom F
16 10 eleq2d F : A B x dom F x A
17 10 eleq2d F : A B y dom F y A
18 16 17 anbi12d F : A B x dom F y dom F x A y A
19 1 2 18 3syl F Isom E , E A B x dom F y dom F x A y A
20 isorel F Isom E , E A B x A y A x E y F x E F y
21 epel x E y x y
22 fvex F y V
23 22 epeli F x E F y F x F y
24 20 21 23 3bitr3g F Isom E , E A B x A y A x y F x F y
25 24 biimpd F Isom E , E A B x A y A x y F x F y
26 25 ex F Isom E , E A B x A y A x y F x F y
27 19 26 sylbid F Isom E , E A B x dom F y dom F x y F x F y
28 27 ralrimivv F Isom E , E A B x dom F y dom F x y F x F y
29 28 3ad2ant1 F Isom E , E A B Ord A B On x dom F y dom F x y F x F y
30 df-smo Smo F F : dom F On Ord dom F x dom F y dom F x y F x F y
31 9 15 29 30 syl3anbrc F Isom E , E A B Ord A B On Smo F