Metamath Proof Explorer


Theorem axcgrtr

Description: Congruence is transitive. Axiom A2 of Schwabhauser p. 10. (Contributed by Scott Fenton, 3-Jun-2013)

Ref Expression
Assertion axcgrtr ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → A B Cgr C D ∧ A B Cgr E F → C D Cgr E F

Proof

Step Hyp Ref Expression
1 eqtr2 ⊢ ∑ i = 1 N A ⁡ i − B ⁡ i 2 = ∑ i = 1 N C ⁡ i − D ⁡ i 2 ∧ ∑ i = 1 N A ⁡ i − B ⁡ i 2 = ∑ i = 1 N E ⁡ i − F ⁡ i 2 → ∑ i = 1 N C ⁡ i − D ⁡ i 2 = ∑ i = 1 N E ⁡ i − F ⁡ i 2
2 simpl1 ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → A ∈ 𝔼 ⁡ N
3 simpl2 ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → B ∈ 𝔼 ⁡ N
4 simpl3 ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → C ∈ 𝔼 ⁡ N
5 simpr1 ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → D ∈ 𝔼 ⁡ N
6 brcgr ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → A B Cgr C D ↔ ∑ i = 1 N A ⁡ i − B ⁡ i 2 = ∑ i = 1 N C ⁡ i − D ⁡ i 2
7 2 3 4 5 6 syl22anc ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → A B Cgr C D ↔ ∑ i = 1 N A ⁡ i − B ⁡ i 2 = ∑ i = 1 N C ⁡ i − D ⁡ i 2
8 simpr2 ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → E ∈ 𝔼 ⁡ N
9 simpr3 ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → F ∈ 𝔼 ⁡ N
10 brcgr ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → A B Cgr E F ↔ ∑ i = 1 N A ⁡ i − B ⁡ i 2 = ∑ i = 1 N E ⁡ i − F ⁡ i 2
11 2 3 8 9 10 syl22anc ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → A B Cgr E F ↔ ∑ i = 1 N A ⁡ i − B ⁡ i 2 = ∑ i = 1 N E ⁡ i − F ⁡ i 2
12 7 11 anbi12d ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → A B Cgr C D ∧ A B Cgr E F ↔ ∑ i = 1 N A ⁡ i − B ⁡ i 2 = ∑ i = 1 N C ⁡ i − D ⁡ i 2 ∧ ∑ i = 1 N A ⁡ i − B ⁡ i 2 = ∑ i = 1 N E ⁡ i − F ⁡ i 2
13 brcgr ⊢ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → C D Cgr E F ↔ ∑ i = 1 N C ⁡ i − D ⁡ i 2 = ∑ i = 1 N E ⁡ i − F ⁡ i 2
14 4 5 8 9 13 syl22anc ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → C D Cgr E F ↔ ∑ i = 1 N C ⁡ i − D ⁡ i 2 = ∑ i = 1 N E ⁡ i − F ⁡ i 2
15 12 14 imbi12d ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → A B Cgr C D ∧ A B Cgr E F → C D Cgr E F ↔ ∑ i = 1 N A ⁡ i − B ⁡ i 2 = ∑ i = 1 N C ⁡ i − D ⁡ i 2 ∧ ∑ i = 1 N A ⁡ i − B ⁡ i 2 = ∑ i = 1 N E ⁡ i − F ⁡ i 2 → ∑ i = 1 N C ⁡ i − D ⁡ i 2 = ∑ i = 1 N E ⁡ i − F ⁡ i 2
16 1 15 mpbiri ⊢ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → A B Cgr C D ∧ A B Cgr E F → C D Cgr E F
17 16 3adant1 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N → A B Cgr C D ∧ A B Cgr E F → C D Cgr E F