This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

Metamath Proof Explorer

Theorem caovdirg

Description: Convert an operation reverse distributive law to class notation. (Contributed by Mario Carneiro, 19-Oct-2014)

Ref Expression
Hypothesis caovdirg.1 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆 ∧ 𝑧 ∈ 𝐾 ) ) → ( ( 𝑥 𝐹 𝑦 ) 𝐺 𝑧 ) = ( ( 𝑥 𝐺 𝑧 ) 𝐻 ( 𝑦 𝐺 𝑧 ) ) )
Assertion caovdirg ( ( 𝜑 ∧ ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ∧ 𝐶 ∈ 𝐾 ) ) → ( ( 𝐴 𝐹 𝐵 ) 𝐺 𝐶 ) = ( ( 𝐴 𝐺 𝐶 ) 𝐻 ( 𝐵 𝐺 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 caovdirg.1 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆 ∧ 𝑧 ∈ 𝐾 ) ) → ( ( 𝑥 𝐹 𝑦 ) 𝐺 𝑧 ) = ( ( 𝑥 𝐺 𝑧 ) 𝐻 ( 𝑦 𝐺 𝑧 ) ) )
2 1 ralrimivvva ⊢ ( 𝜑 → ∀ 𝑥 ∈ 𝑆 ∀ 𝑦 ∈ 𝑆 ∀ 𝑧 ∈ 𝐾 ( ( 𝑥 𝐹 𝑦 ) 𝐺 𝑧 ) = ( ( 𝑥 𝐺 𝑧 ) 𝐻 ( 𝑦 𝐺 𝑧 ) ) )
3 oveq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 𝐹 𝑦 ) = ( 𝐴 𝐹 𝑦 ) )
4 3 oveq1d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 𝐹 𝑦 ) 𝐺 𝑧 ) = ( ( 𝐴 𝐹 𝑦 ) 𝐺 𝑧 ) )
5 oveq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 𝐺 𝑧 ) = ( 𝐴 𝐺 𝑧 ) )
6 5 oveq1d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 𝐺 𝑧 ) 𝐻 ( 𝑦 𝐺 𝑧 ) ) = ( ( 𝐴 𝐺 𝑧 ) 𝐻 ( 𝑦 𝐺 𝑧 ) ) )
7 4 6 eqeq12d ⊢ ( 𝑥 = 𝐴 → ( ( ( 𝑥 𝐹 𝑦 ) 𝐺 𝑧 ) = ( ( 𝑥 𝐺 𝑧 ) 𝐻 ( 𝑦 𝐺 𝑧 ) ) ↔ ( ( 𝐴 𝐹 𝑦 ) 𝐺 𝑧 ) = ( ( 𝐴 𝐺 𝑧 ) 𝐻 ( 𝑦 𝐺 𝑧 ) ) ) )
8 oveq2 ⊢ ( 𝑦 = 𝐵 → ( 𝐴 𝐹 𝑦 ) = ( 𝐴 𝐹 𝐵 ) )
9 8 oveq1d ⊢ ( 𝑦 = 𝐵 → ( ( 𝐴 𝐹 𝑦 ) 𝐺 𝑧 ) = ( ( 𝐴 𝐹 𝐵 ) 𝐺 𝑧 ) )
10 oveq1 ⊢ ( 𝑦 = 𝐵 → ( 𝑦 𝐺 𝑧 ) = ( 𝐵 𝐺 𝑧 ) )
11 10 oveq2d ⊢ ( 𝑦 = 𝐵 → ( ( 𝐴 𝐺 𝑧 ) 𝐻 ( 𝑦 𝐺 𝑧 ) ) = ( ( 𝐴 𝐺 𝑧 ) 𝐻 ( 𝐵 𝐺 𝑧 ) ) )
12 9 11 eqeq12d ⊢ ( 𝑦 = 𝐵 → ( ( ( 𝐴 𝐹 𝑦 ) 𝐺 𝑧 ) = ( ( 𝐴 𝐺 𝑧 ) 𝐻 ( 𝑦 𝐺 𝑧 ) ) ↔ ( ( 𝐴 𝐹 𝐵 ) 𝐺 𝑧 ) = ( ( 𝐴 𝐺 𝑧 ) 𝐻 ( 𝐵 𝐺 𝑧 ) ) ) )
13 oveq2 ⊢ ( 𝑧 = 𝐶 → ( ( 𝐴 𝐹 𝐵 ) 𝐺 𝑧 ) = ( ( 𝐴 𝐹 𝐵 ) 𝐺 𝐶 ) )
14 oveq2 ⊢ ( 𝑧 = 𝐶 → ( 𝐴 𝐺 𝑧 ) = ( 𝐴 𝐺 𝐶 ) )
15 oveq2 ⊢ ( 𝑧 = 𝐶 → ( 𝐵 𝐺 𝑧 ) = ( 𝐵 𝐺 𝐶 ) )
16 14 15 oveq12d ⊢ ( 𝑧 = 𝐶 → ( ( 𝐴 𝐺 𝑧 ) 𝐻 ( 𝐵 𝐺 𝑧 ) ) = ( ( 𝐴 𝐺 𝐶 ) 𝐻 ( 𝐵 𝐺 𝐶 ) ) )
17 13 16 eqeq12d ⊢ ( 𝑧 = 𝐶 → ( ( ( 𝐴 𝐹 𝐵 ) 𝐺 𝑧 ) = ( ( 𝐴 𝐺 𝑧 ) 𝐻 ( 𝐵 𝐺 𝑧 ) ) ↔ ( ( 𝐴 𝐹 𝐵 ) 𝐺 𝐶 ) = ( ( 𝐴 𝐺 𝐶 ) 𝐻 ( 𝐵 𝐺 𝐶 ) ) ) )
18 7 12 17 rspc3v ⊢ ( ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ∧ 𝐶 ∈ 𝐾 ) → ( ∀ 𝑥 ∈ 𝑆 ∀ 𝑦 ∈ 𝑆 ∀ 𝑧 ∈ 𝐾 ( ( 𝑥 𝐹 𝑦 ) 𝐺 𝑧 ) = ( ( 𝑥 𝐺 𝑧 ) 𝐻 ( 𝑦 𝐺 𝑧 ) ) → ( ( 𝐴 𝐹 𝐵 ) 𝐺 𝐶 ) = ( ( 𝐴 𝐺 𝐶 ) 𝐻 ( 𝐵 𝐺 𝐶 ) ) ) )
19 2 18 mpan9 ⊢ ( ( 𝜑 ∧ ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ∧ 𝐶 ∈ 𝐾 ) ) → ( ( 𝐴 𝐹 𝐵 ) 𝐺 𝐶 ) = ( ( 𝐴 𝐺 𝐶 ) 𝐻 ( 𝐵 𝐺 𝐶 ) ) )