Metamath Proof Explorer


Theorem wemoiso

Description: Thus, there is at most one isomorphism between any two well-ordered sets. TODO: Shorten finnisoeu . (Contributed by Stefan O'Rear, 12-Feb-2015) (Revised by Mario Carneiro, 25-Jun-2015)

Ref Expression
Assertion wemoiso ( 𝑅 We 𝐴 → ∃* 𝑓 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) )

Proof

Step Hyp Ref Expression
1 simpl ( ( 𝑅 We 𝐴 ∧ ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ∧ 𝑔 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ) ) → 𝑅 We 𝐴 )
2 vex 𝑓 ∈ V
3 isof1o ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) → 𝑓 : 𝐴1-1-onto𝐵 )
4 f1of ( 𝑓 : 𝐴1-1-onto𝐵𝑓 : 𝐴𝐵 )
5 3 4 syl ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) → 𝑓 : 𝐴𝐵 )
6 dmfex ( ( 𝑓 ∈ V ∧ 𝑓 : 𝐴𝐵 ) → 𝐴 ∈ V )
7 2 5 6 sylancr ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) → 𝐴 ∈ V )
8 7 ad2antrl ( ( 𝑅 We 𝐴 ∧ ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ∧ 𝑔 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ) ) → 𝐴 ∈ V )
9 exse ( 𝐴 ∈ V → 𝑅 Se 𝐴 )
10 8 9 syl ( ( 𝑅 We 𝐴 ∧ ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ∧ 𝑔 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ) ) → 𝑅 Se 𝐴 )
11 1 10 jca ( ( 𝑅 We 𝐴 ∧ ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ∧ 𝑔 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ) ) → ( 𝑅 We 𝐴𝑅 Se 𝐴 ) )
12 weisoeq ( ( ( 𝑅 We 𝐴𝑅 Se 𝐴 ) ∧ ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ∧ 𝑔 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ) ) → 𝑓 = 𝑔 )
13 11 12 sylancom ( ( 𝑅 We 𝐴 ∧ ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ∧ 𝑔 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ) ) → 𝑓 = 𝑔 )
14 13 ex ( 𝑅 We 𝐴 → ( ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ∧ 𝑔 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ) → 𝑓 = 𝑔 ) )
15 14 alrimivv ( 𝑅 We 𝐴 → ∀ 𝑓𝑔 ( ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ∧ 𝑔 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ) → 𝑓 = 𝑔 ) )
16 isoeq1 ( 𝑓 = 𝑔 → ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ↔ 𝑔 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ) )
17 16 mo4 ( ∃* 𝑓 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ↔ ∀ 𝑓𝑔 ( ( 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ∧ 𝑔 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ) → 𝑓 = 𝑔 ) )
18 15 17 sylibr ( 𝑅 We 𝐴 → ∃* 𝑓 𝑓 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) )