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 simpl ( ( 𝐹 Fn 𝑋𝑌𝑋 ) → 𝐹 Fn 𝑋 )
5 dffn2 ( 𝐹 Fn 𝑋𝐹 : 𝑋 ⟶ V )
6 4 5 sylib ( ( 𝐹 Fn 𝑋𝑌𝑋 ) → 𝐹 : 𝑋 ⟶ V )
7 6 feqmptd ( ( 𝐹 Fn 𝑋𝑌𝑋 ) → 𝐹 = ( 𝑦𝑋 ↦ ( 𝐹𝑦 ) ) )
8 fveq2 ( 𝑦 = 𝑌 → ( 𝐹𝑦 ) = ( 𝐹𝑌 ) )
9 1 3 7 8 fmptco ( ( 𝐹 Fn 𝑋𝑌𝑋 ) → ( 𝐹 ∘ ( 𝐼 × { 𝑌 } ) ) = ( 𝑥𝐼 ↦ ( 𝐹𝑌 ) ) )
10 fconstmpt ( 𝐼 × { ( 𝐹𝑌 ) } ) = ( 𝑥𝐼 ↦ ( 𝐹𝑌 ) )
11 9 10 eqtr4di ( ( 𝐹 Fn 𝑋𝑌𝑋 ) → ( 𝐹 ∘ ( 𝐼 × { 𝑌 } ) ) = ( 𝐼 × { ( 𝐹𝑌 ) } ) )