Metamath Proof Explorer


Theorem fconst6g

Description: Constant function with loose range. (Contributed by Stefan O'Rear, 1-Feb-2015)

Ref Expression
Assertion fconst6g ( 𝐵𝐶 → ( 𝐴 × { 𝐵 } ) : 𝐴𝐶 )

Proof

Step Hyp Ref Expression
1 fconstg ( 𝐵𝐶 → ( 𝐴 × { 𝐵 } ) : 𝐴 ⟶ { 𝐵 } )
2 snssi ( 𝐵𝐶 → { 𝐵 } ⊆ 𝐶 )
3 1 2 fssd ( 𝐵𝐶 → ( 𝐴 × { 𝐵 } ) : 𝐴𝐶 )