Metamath Proof Explorer


Theorem dvgt0lem2

Description: Lemma for dvgt0 and dvlt0 . (Contributed by Mario Carneiro, 19-Feb-2015)

Ref Expression
Hypotheses dvgt0.a φ A
dvgt0.b φ B
dvgt0.f φ F : A B cn
dvgt0lem.d φ F : A B S
dvgt0lem.o O Or
dvgt0lem.i φ x A B y A B x < y F x O F y
Assertion dvgt0lem2 φ F Isom < , O A B ran F

Proof

Step Hyp Ref Expression
1 dvgt0.a φ A
2 dvgt0.b φ B
3 dvgt0.f φ F : A B cn
4 dvgt0lem.d φ F : A B S
5 dvgt0lem.o O Or
6 dvgt0lem.i φ x A B y A B x < y F x O F y
7 6 ex φ x A B y A B x < y F x O F y
8 7 ralrimivva φ x A B y A B x < y F x O F y
9 iccssre A B A B
10 1 2 9 syl2anc φ A B
11 ltso < Or
12 soss A B < Or < Or A B
13 10 11 12 mpisyl φ < Or A B
14 5 a1i φ O Or
15 cncff F : A B cn F : A B
16 3 15 syl φ F : A B
17 ssidd φ A B A B
18 soisores < Or A B O Or F : A B A B A B F A B Isom < , O A B F A B x A B y A B x < y F x O F y
19 13 14 16 17 18 syl22anc φ F A B Isom < , O A B F A B x A B y A B x < y F x O F y
20 8 19 mpbird φ F A B Isom < , O A B F A B
21 ffn F : A B F Fn A B
22 3 15 21 3syl φ F Fn A B
23 fnresdm F Fn A B F A B = F
24 isoeq1 F A B = F F A B Isom < , O A B F A B F Isom < , O A B F A B
25 22 23 24 3syl φ F A B Isom < , O A B F A B F Isom < , O A B F A B
26 20 25 mpbid φ F Isom < , O A B F A B
27 fnima F Fn A B F A B = ran F
28 isoeq5 F A B = ran F F Isom < , O A B F A B F Isom < , O A B ran F
29 22 27 28 3syl φ F Isom < , O A B F A B F Isom < , O A B ran F
30 26 29 mpbid φ F Isom < , O A B ran F