Metamath Proof Explorer


Theorem 2cycl2d

Description: Construction of a 2-cycle from two given edges in a graph. (Contributed by BTernaryTau, 16-Oct-2023)

Ref Expression
Hypotheses 2cycl2d.1 𝑃 = ⟨“ 𝐴 𝐵 𝐴 ”⟩
2cycl2d.2 𝐹 = ⟨“ 𝐽 𝐾 ”⟩
2cycl2d.3 ( 𝜑 → ( 𝐴𝑉𝐵𝑉 ) )
2cycl2d.4 ( 𝜑𝐴𝐵 )
2cycl2d.5 ( 𝜑 → ( { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐽 ) ∧ { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐾 ) ) )
2cycl2d.6 𝑉 = ( Vtx ‘ 𝐺 )
2cycl2d.7 𝐼 = ( iEdg ‘ 𝐺 )
2cycl2d.8 ( 𝜑𝐽𝐾 )
Assertion 2cycl2d ( 𝜑𝐹 ( Cycles ‘ 𝐺 ) 𝑃 )

Proof

Step Hyp Ref Expression
1 2cycl2d.1 𝑃 = ⟨“ 𝐴 𝐵 𝐴 ”⟩
2 2cycl2d.2 𝐹 = ⟨“ 𝐽 𝐾 ”⟩
3 2cycl2d.3 ( 𝜑 → ( 𝐴𝑉𝐵𝑉 ) )
4 2cycl2d.4 ( 𝜑𝐴𝐵 )
5 2cycl2d.5 ( 𝜑 → ( { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐽 ) ∧ { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐾 ) ) )
6 2cycl2d.6 𝑉 = ( Vtx ‘ 𝐺 )
7 2cycl2d.7 𝐼 = ( iEdg ‘ 𝐺 )
8 2cycl2d.8 ( 𝜑𝐽𝐾 )
9 simpl ( ( 𝐴𝑉𝐵𝑉 ) → 𝐴𝑉 )
10 3 9 jccir ( 𝜑 → ( ( 𝐴𝑉𝐵𝑉 ) ∧ 𝐴𝑉 ) )
11 df-3an ( ( 𝐴𝑉𝐵𝑉𝐴𝑉 ) ↔ ( ( 𝐴𝑉𝐵𝑉 ) ∧ 𝐴𝑉 ) )
12 10 11 sylibr ( 𝜑 → ( 𝐴𝑉𝐵𝑉𝐴𝑉 ) )
13 4 necomd ( 𝜑𝐵𝐴 )
14 4 13 jca ( 𝜑 → ( 𝐴𝐵𝐵𝐴 ) )
15 prcom { 𝐴 , 𝐵 } = { 𝐵 , 𝐴 }
16 15 sseq1i ( { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐾 ) ↔ { 𝐵 , 𝐴 } ⊆ ( 𝐼𝐾 ) )
17 16 anbi2i ( ( { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐽 ) ∧ { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐾 ) ) ↔ ( { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐽 ) ∧ { 𝐵 , 𝐴 } ⊆ ( 𝐼𝐾 ) ) )
18 5 17 sylib ( 𝜑 → ( { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐽 ) ∧ { 𝐵 , 𝐴 } ⊆ ( 𝐼𝐾 ) ) )
19 eqidd ( 𝜑𝐴 = 𝐴 )
20 1 2 12 14 18 6 7 8 19 2cycld ( 𝜑𝐹 ( Cycles ‘ 𝐺 ) 𝑃 )