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Metamath Proof Explorer

Theorem caovordg

Description: Convert an operation ordering law to class notation. (Contributed by NM, 19-Feb-1996) (Revised by Mario Carneiro, 30-Dec-2014)

Ref Expression
Hypothesis caovordg.1 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆 ∧ 𝑧 ∈ 𝑆 ) ) → ( 𝑥 𝑅 𝑦 ↔ ( 𝑧 𝐹 𝑥 ) 𝑅 ( 𝑧 𝐹 𝑦 ) ) )
Assertion caovordg ( ( 𝜑 ∧ ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ∧ 𝐶 ∈ 𝑆 ) ) → ( 𝐴 𝑅 𝐵 ↔ ( 𝐶 𝐹 𝐴 ) 𝑅 ( 𝐶 𝐹 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 caovordg.1 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆 ∧ 𝑧 ∈ 𝑆 ) ) → ( 𝑥 𝑅 𝑦 ↔ ( 𝑧 𝐹 𝑥 ) 𝑅 ( 𝑧 𝐹 𝑦 ) ) )
2 1 ralrimivvva ⊢ ( 𝜑 → ∀ 𝑥 ∈ 𝑆 ∀ 𝑦 ∈ 𝑆 ∀ 𝑧 ∈ 𝑆 ( 𝑥 𝑅 𝑦 ↔ ( 𝑧 𝐹 𝑥 ) 𝑅 ( 𝑧 𝐹 𝑦 ) ) )
3 breq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 𝑅 𝑦 ↔ 𝐴 𝑅 𝑦 ) )
4 oveq2 ⊢ ( 𝑥 = 𝐴 → ( 𝑧 𝐹 𝑥 ) = ( 𝑧 𝐹 𝐴 ) )
5 4 breq1d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑧 𝐹 𝑥 ) 𝑅 ( 𝑧 𝐹 𝑦 ) ↔ ( 𝑧 𝐹 𝐴 ) 𝑅 ( 𝑧 𝐹 𝑦 ) ) )
6 3 5 bibi12d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 𝑅 𝑦 ↔ ( 𝑧 𝐹 𝑥 ) 𝑅 ( 𝑧 𝐹 𝑦 ) ) ↔ ( 𝐴 𝑅 𝑦 ↔ ( 𝑧 𝐹 𝐴 ) 𝑅 ( 𝑧 𝐹 𝑦 ) ) ) )
7 breq2 ⊢ ( 𝑦 = 𝐵 → ( 𝐴 𝑅 𝑦 ↔ 𝐴 𝑅 𝐵 ) )
8 oveq2 ⊢ ( 𝑦 = 𝐵 → ( 𝑧 𝐹 𝑦 ) = ( 𝑧 𝐹 𝐵 ) )
9 8 breq2d ⊢ ( 𝑦 = 𝐵 → ( ( 𝑧 𝐹 𝐴 ) 𝑅 ( 𝑧 𝐹 𝑦 ) ↔ ( 𝑧 𝐹 𝐴 ) 𝑅 ( 𝑧 𝐹 𝐵 ) ) )
10 7 9 bibi12d ⊢ ( 𝑦 = 𝐵 → ( ( 𝐴 𝑅 𝑦 ↔ ( 𝑧 𝐹 𝐴 ) 𝑅 ( 𝑧 𝐹 𝑦 ) ) ↔ ( 𝐴 𝑅 𝐵 ↔ ( 𝑧 𝐹 𝐴 ) 𝑅 ( 𝑧 𝐹 𝐵 ) ) ) )
11 oveq1 ⊢ ( 𝑧 = 𝐶 → ( 𝑧 𝐹 𝐴 ) = ( 𝐶 𝐹 𝐴 ) )
12 oveq1 ⊢ ( 𝑧 = 𝐶 → ( 𝑧 𝐹 𝐵 ) = ( 𝐶 𝐹 𝐵 ) )
13 11 12 breq12d ⊢ ( 𝑧 = 𝐶 → ( ( 𝑧 𝐹 𝐴 ) 𝑅 ( 𝑧 𝐹 𝐵 ) ↔ ( 𝐶 𝐹 𝐴 ) 𝑅 ( 𝐶 𝐹 𝐵 ) ) )
14 13 bibi2d ⊢ ( 𝑧 = 𝐶 → ( ( 𝐴 𝑅 𝐵 ↔ ( 𝑧 𝐹 𝐴 ) 𝑅 ( 𝑧 𝐹 𝐵 ) ) ↔ ( 𝐴 𝑅 𝐵 ↔ ( 𝐶 𝐹 𝐴 ) 𝑅 ( 𝐶 𝐹 𝐵 ) ) ) )
15 6 10 14 rspc3v ⊢ ( ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ∧ 𝐶 ∈ 𝑆 ) → ( ∀ 𝑥 ∈ 𝑆 ∀ 𝑦 ∈ 𝑆 ∀ 𝑧 ∈ 𝑆 ( 𝑥 𝑅 𝑦 ↔ ( 𝑧 𝐹 𝑥 ) 𝑅 ( 𝑧 𝐹 𝑦 ) ) → ( 𝐴 𝑅 𝐵 ↔ ( 𝐶 𝐹 𝐴 ) 𝑅 ( 𝐶 𝐹 𝐵 ) ) ) )
16 2 15 mpan9 ⊢ ( ( 𝜑 ∧ ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ∧ 𝐶 ∈ 𝑆 ) ) → ( 𝐴 𝑅 𝐵 ↔ ( 𝐶 𝐹 𝐴 ) 𝑅 ( 𝐶 𝐹 𝐵 ) ) )