Metamath Proof Explorer


Theorem gpgprismgr4cycllem1

Description: Lemma 1 for gpgprismgr4cycl0 : the cycle <. P , F >. consists of 4 edges (i.e., has length 4). (Contributed by AV, 1-Nov-2025)

Ref Expression
Hypothesis gpgprismgr4cycllem1.f 𝐹 = ⟨“ { ⟨ 0 , 0 ⟩ , ⟨ 0 , 1 ⟩ } { ⟨ 0 , 1 ⟩ , ⟨ 1 , 1 ⟩ } { ⟨ 1 , 1 ⟩ , ⟨ 1 , 0 ⟩ } { ⟨ 1 , 0 ⟩ , ⟨ 0 , 0 ⟩ } ”⟩
Assertion gpgprismgr4cycllem1 ( ♯ ‘ 𝐹 ) = 4

Proof

Step Hyp Ref Expression
1 gpgprismgr4cycllem1.f 𝐹 = ⟨“ { ⟨ 0 , 0 ⟩ , ⟨ 0 , 1 ⟩ } { ⟨ 0 , 1 ⟩ , ⟨ 1 , 1 ⟩ } { ⟨ 1 , 1 ⟩ , ⟨ 1 , 0 ⟩ } { ⟨ 1 , 0 ⟩ , ⟨ 0 , 0 ⟩ } ”⟩
2 1 fveq2i ( ♯ ‘ 𝐹 ) = ( ♯ ‘ ⟨“ { ⟨ 0 , 0 ⟩ , ⟨ 0 , 1 ⟩ } { ⟨ 0 , 1 ⟩ , ⟨ 1 , 1 ⟩ } { ⟨ 1 , 1 ⟩ , ⟨ 1 , 0 ⟩ } { ⟨ 1 , 0 ⟩ , ⟨ 0 , 0 ⟩ } ”⟩ )
3 s4len ( ♯ ‘ ⟨“ { ⟨ 0 , 0 ⟩ , ⟨ 0 , 1 ⟩ } { ⟨ 0 , 1 ⟩ , ⟨ 1 , 1 ⟩ } { ⟨ 1 , 1 ⟩ , ⟨ 1 , 0 ⟩ } { ⟨ 1 , 0 ⟩ , ⟨ 0 , 0 ⟩ } ”⟩ ) = 4
4 2 3 eqtri ( ♯ ‘ 𝐹 ) = 4