Metamath Proof Explorer


Theorem gictr

Description: Isomorphism is transitive. (Contributed by Mario Carneiro, 21-Apr-2016)

Ref Expression
Assertion gictr ( ( 𝑅𝑔 𝑆𝑆𝑔 𝑇 ) → 𝑅𝑔 𝑇 )

Proof

Step Hyp Ref Expression
1 brgic ( 𝑅𝑔 𝑆 ↔ ( 𝑅 GrpIso 𝑆 ) ≠ ∅ )
2 brgic ( 𝑆𝑔 𝑇 ↔ ( 𝑆 GrpIso 𝑇 ) ≠ ∅ )
3 n0 ( ( 𝑅 GrpIso 𝑆 ) ≠ ∅ ↔ ∃ 𝑓 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) )
4 n0 ( ( 𝑆 GrpIso 𝑇 ) ≠ ∅ ↔ ∃ 𝑔 𝑔 ∈ ( 𝑆 GrpIso 𝑇 ) )
5 exdistrv ( ∃ 𝑓𝑔 ( 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) ∧ 𝑔 ∈ ( 𝑆 GrpIso 𝑇 ) ) ↔ ( ∃ 𝑓 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) ∧ ∃ 𝑔 𝑔 ∈ ( 𝑆 GrpIso 𝑇 ) ) )
6 gimco ( ( 𝑔 ∈ ( 𝑆 GrpIso 𝑇 ) ∧ 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) ) → ( 𝑔𝑓 ) ∈ ( 𝑅 GrpIso 𝑇 ) )
7 brgici ( ( 𝑔𝑓 ) ∈ ( 𝑅 GrpIso 𝑇 ) → 𝑅𝑔 𝑇 )
8 6 7 syl ( ( 𝑔 ∈ ( 𝑆 GrpIso 𝑇 ) ∧ 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) ) → 𝑅𝑔 𝑇 )
9 8 ancoms ( ( 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) ∧ 𝑔 ∈ ( 𝑆 GrpIso 𝑇 ) ) → 𝑅𝑔 𝑇 )
10 9 exlimivv ( ∃ 𝑓𝑔 ( 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) ∧ 𝑔 ∈ ( 𝑆 GrpIso 𝑇 ) ) → 𝑅𝑔 𝑇 )
11 5 10 sylbir ( ( ∃ 𝑓 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) ∧ ∃ 𝑔 𝑔 ∈ ( 𝑆 GrpIso 𝑇 ) ) → 𝑅𝑔 𝑇 )
12 3 4 11 syl2anb ( ( ( 𝑅 GrpIso 𝑆 ) ≠ ∅ ∧ ( 𝑆 GrpIso 𝑇 ) ≠ ∅ ) → 𝑅𝑔 𝑇 )
13 1 2 12 syl2anb ( ( 𝑅𝑔 𝑆𝑆𝑔 𝑇 ) → 𝑅𝑔 𝑇 )