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Metamath Proof Explorer

Theorem caovcang

Description: Convert an operation cancellation law to class notation. (Contributed by NM, 20-Aug-1995) (Revised by Mario Carneiro, 30-Dec-2014)

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

Proof

Step Hyp Ref Expression
1 caovcang.1 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝑇 ∧ 𝑦 ∈ 𝑆 ∧ 𝑧 ∈ 𝑆 ) ) → ( ( 𝑥 𝐹 𝑦 ) = ( 𝑥 𝐹 𝑧 ) ↔ 𝑦 = 𝑧 ) )
2 1 ralrimivvva ⊢ ( 𝜑 → ∀ 𝑥 ∈ 𝑇 ∀ 𝑦 ∈ 𝑆 ∀ 𝑧 ∈ 𝑆 ( ( 𝑥 𝐹 𝑦 ) = ( 𝑥 𝐹 𝑧 ) ↔ 𝑦 = 𝑧 ) )
3 oveq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 𝐹 𝑦 ) = ( 𝐴 𝐹 𝑦 ) )
4 oveq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 𝐹 𝑧 ) = ( 𝐴 𝐹 𝑧 ) )
5 3 4 eqeq12d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 𝐹 𝑦 ) = ( 𝑥 𝐹 𝑧 ) ↔ ( 𝐴 𝐹 𝑦 ) = ( 𝐴 𝐹 𝑧 ) ) )
6 5 bibi1d ⊢ ( 𝑥 = 𝐴 → ( ( ( 𝑥 𝐹 𝑦 ) = ( 𝑥 𝐹 𝑧 ) ↔ 𝑦 = 𝑧 ) ↔ ( ( 𝐴 𝐹 𝑦 ) = ( 𝐴 𝐹 𝑧 ) ↔ 𝑦 = 𝑧 ) ) )
7 oveq2 ⊢ ( 𝑦 = 𝐵 → ( 𝐴 𝐹 𝑦 ) = ( 𝐴 𝐹 𝐵 ) )
8 7 eqeq1d ⊢ ( 𝑦 = 𝐵 → ( ( 𝐴 𝐹 𝑦 ) = ( 𝐴 𝐹 𝑧 ) ↔ ( 𝐴 𝐹 𝐵 ) = ( 𝐴 𝐹 𝑧 ) ) )
9 eqeq1 ⊢ ( 𝑦 = 𝐵 → ( 𝑦 = 𝑧 ↔ 𝐵 = 𝑧 ) )
10 8 9 bibi12d ⊢ ( 𝑦 = 𝐵 → ( ( ( 𝐴 𝐹 𝑦 ) = ( 𝐴 𝐹 𝑧 ) ↔ 𝑦 = 𝑧 ) ↔ ( ( 𝐴 𝐹 𝐵 ) = ( 𝐴 𝐹 𝑧 ) ↔ 𝐵 = 𝑧 ) ) )
11 oveq2 ⊢ ( 𝑧 = 𝐶 → ( 𝐴 𝐹 𝑧 ) = ( 𝐴 𝐹 𝐶 ) )
12 11 eqeq2d ⊢ ( 𝑧 = 𝐶 → ( ( 𝐴 𝐹 𝐵 ) = ( 𝐴 𝐹 𝑧 ) ↔ ( 𝐴 𝐹 𝐵 ) = ( 𝐴 𝐹 𝐶 ) ) )
13 eqeq2 ⊢ ( 𝑧 = 𝐶 → ( 𝐵 = 𝑧 ↔ 𝐵 = 𝐶 ) )
14 12 13 bibi12d ⊢ ( 𝑧 = 𝐶 → ( ( ( 𝐴 𝐹 𝐵 ) = ( 𝐴 𝐹 𝑧 ) ↔ 𝐵 = 𝑧 ) ↔ ( ( 𝐴 𝐹 𝐵 ) = ( 𝐴 𝐹 𝐶 ) ↔ 𝐵 = 𝐶 ) ) )
15 6 10 14 rspc3v ⊢ ( ( 𝐴 ∈ 𝑇 ∧ 𝐵 ∈ 𝑆 ∧ 𝐶 ∈ 𝑆 ) → ( ∀ 𝑥 ∈ 𝑇 ∀ 𝑦 ∈ 𝑆 ∀ 𝑧 ∈ 𝑆 ( ( 𝑥 𝐹 𝑦 ) = ( 𝑥 𝐹 𝑧 ) ↔ 𝑦 = 𝑧 ) → ( ( 𝐴 𝐹 𝐵 ) = ( 𝐴 𝐹 𝐶 ) ↔ 𝐵 = 𝐶 ) ) )
16 2 15 mpan9 ⊢ ( ( 𝜑 ∧ ( 𝐴 ∈ 𝑇 ∧ 𝐵 ∈ 𝑆 ∧ 𝐶 ∈ 𝑆 ) ) → ( ( 𝐴 𝐹 𝐵 ) = ( 𝐴 𝐹 𝐶 ) ↔ 𝐵 = 𝐶 ) )