Metamath Proof Explorer


Theorem fcoconst

Description: Composition with a constant function. (Contributed by Stefan O'Rear, 11-Mar-2015)

Ref Expression
Assertion fcoconst ( ( 𝐹 Fn 𝑋𝑌𝑋 ) → ( 𝐹 ∘ ( 𝐼 × { 𝑌 } ) ) = ( 𝐼 × { ( 𝐹𝑌 ) } ) )

Proof

Step Hyp Ref Expression
1 simplr ( ( ( 𝐹 Fn 𝑋𝑌𝑋 ) ∧ 𝑥𝐼 ) → 𝑌𝑋 )
2 fconstmpt ( 𝐼 × { 𝑌 } ) = ( 𝑥𝐼𝑌 )
3 2 a1i ( ( 𝐹 Fn 𝑋𝑌𝑋 ) → ( 𝐼 × { 𝑌 } ) = ( 𝑥𝐼𝑌 ) )
4 dffn2 ( 𝐹 Fn 𝑋𝐹 : 𝑋 ⟶ V )
5 4 birani ( ( 𝐹 Fn 𝑋𝑌𝑋 ) → 𝐹 : 𝑋 ⟶ V )
6 5 feqmptd ( ( 𝐹 Fn 𝑋𝑌𝑋 ) → 𝐹 = ( 𝑦𝑋 ↦ ( 𝐹𝑦 ) ) )
7 fveq2 ( 𝑦 = 𝑌 → ( 𝐹𝑦 ) = ( 𝐹𝑌 ) )
8 1 3 6 7 fmptco ( ( 𝐹 Fn 𝑋𝑌𝑋 ) → ( 𝐹 ∘ ( 𝐼 × { 𝑌 } ) ) = ( 𝑥𝐼 ↦ ( 𝐹𝑌 ) ) )
9 fconstmpt ( 𝐼 × { ( 𝐹𝑌 ) } ) = ( 𝑥𝐼 ↦ ( 𝐹𝑌 ) )
10 8 9 eqtr4di ( ( 𝐹 Fn 𝑋𝑌𝑋 ) → ( 𝐹 ∘ ( 𝐼 × { 𝑌 } ) ) = ( 𝐼 × { ( 𝐹𝑌 ) } ) )