Metamath Proof Explorer


Theorem ofcfeqd2

Description: Equality theorem for function/constant operation value. (Contributed by Thierry Arnoux, 31-Jan-2017)

Ref Expression
Hypotheses ofcfeqd2.1 ( ( 𝜑𝑥𝐴 ) → ( 𝐹𝑥 ) ∈ 𝐵 )
ofcfeqd2.2 ( ( 𝜑𝑦𝐵 ) → ( 𝑦 𝑅 𝐶 ) = ( 𝑦 𝑃 𝐶 ) )
ofcfeqd2.3 ( 𝜑𝐹 Fn 𝐴 )
ofcfeqd2.4 ( 𝜑𝐴𝑉 )
ofcfeqd2.5 ( 𝜑𝐶𝑊 )
Assertion ofcfeqd2 ( 𝜑 → ( 𝐹f/c 𝑅 𝐶 ) = ( 𝐹f/c 𝑃 𝐶 ) )

Proof

Step Hyp Ref Expression
1 ofcfeqd2.1 ( ( 𝜑𝑥𝐴 ) → ( 𝐹𝑥 ) ∈ 𝐵 )
2 ofcfeqd2.2 ( ( 𝜑𝑦𝐵 ) → ( 𝑦 𝑅 𝐶 ) = ( 𝑦 𝑃 𝐶 ) )
3 ofcfeqd2.3 ( 𝜑𝐹 Fn 𝐴 )
4 ofcfeqd2.4 ( 𝜑𝐴𝑉 )
5 ofcfeqd2.5 ( 𝜑𝐶𝑊 )
6 oveq1 ( 𝑦 = ( 𝐹𝑥 ) → ( 𝑦 𝑅 𝐶 ) = ( ( 𝐹𝑥 ) 𝑅 𝐶 ) )
7 oveq1 ( 𝑦 = ( 𝐹𝑥 ) → ( 𝑦 𝑃 𝐶 ) = ( ( 𝐹𝑥 ) 𝑃 𝐶 ) )
8 6 7 eqeq12d ( 𝑦 = ( 𝐹𝑥 ) → ( ( 𝑦 𝑅 𝐶 ) = ( 𝑦 𝑃 𝐶 ) ↔ ( ( 𝐹𝑥 ) 𝑅 𝐶 ) = ( ( 𝐹𝑥 ) 𝑃 𝐶 ) ) )
9 2 ralrimiva ( 𝜑 → ∀ 𝑦𝐵 ( 𝑦 𝑅 𝐶 ) = ( 𝑦 𝑃 𝐶 ) )
10 9 adantr ( ( 𝜑𝑥𝐴 ) → ∀ 𝑦𝐵 ( 𝑦 𝑅 𝐶 ) = ( 𝑦 𝑃 𝐶 ) )
11 8 10 1 rspcdva ( ( 𝜑𝑥𝐴 ) → ( ( 𝐹𝑥 ) 𝑅 𝐶 ) = ( ( 𝐹𝑥 ) 𝑃 𝐶 ) )
12 11 mpteq2dva ( 𝜑 → ( 𝑥𝐴 ↦ ( ( 𝐹𝑥 ) 𝑅 𝐶 ) ) = ( 𝑥𝐴 ↦ ( ( 𝐹𝑥 ) 𝑃 𝐶 ) ) )
13 eqidd ( ( 𝜑𝑥𝐴 ) → ( 𝐹𝑥 ) = ( 𝐹𝑥 ) )
14 3 4 5 13 ofcfval ( 𝜑 → ( 𝐹f/c 𝑅 𝐶 ) = ( 𝑥𝐴 ↦ ( ( 𝐹𝑥 ) 𝑅 𝐶 ) ) )
15 3 4 5 13 ofcfval ( 𝜑 → ( 𝐹f/c 𝑃 𝐶 ) = ( 𝑥𝐴 ↦ ( ( 𝐹𝑥 ) 𝑃 𝐶 ) ) )
16 12 14 15 3eqtr4d ( 𝜑 → ( 𝐹f/c 𝑅 𝐶 ) = ( 𝐹f/c 𝑃 𝐶 ) )