Metamath Proof Explorer

Theorem cgr3permute5

Description: Permutation law for three-place congruence. (Contributed by Scott Fenton, 5-Oct-2013)

Ref Expression
Assertion cgr3permute5 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → A B C Cgr3 D E F ↔ C B A Cgr3 F E D

Proof

Step Hyp Ref Expression
1 cgr3permute3 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → A B C Cgr3 D E F ↔ B C A Cgr3 E F D
2 biid ⊢ N ∈ ℕ ↔ N ∈ ℕ
3 3anrot ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ↔ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N
4 3anrot ⊢ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N ↔ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N
5 cgr3permute2 ⊢ N ∈ ℕ ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → B C A Cgr3 E F D ↔ C B A Cgr3 F E D
6 2 3 4 5 syl3anb ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → B C A Cgr3 E F D ↔ C B A Cgr3 F E D
7 1 6 bitrd ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → A B C Cgr3 D E F ↔ C B A Cgr3 F E D