Metamath Proof Explorer


Theorem funresdfunsn

Description: Restricting a function to a domain without one element of the domain of the function, and adding a pair of this element and the function value of the element results in the function itself. (Contributed by AV, 2-Dec-2018)

Ref Expression
Assertion funresdfunsn ( ( Fun 𝐹𝑋 ∈ dom 𝐹 ) → ( ( 𝐹 ↾ ( V ∖ { 𝑋 } ) ) ∪ { ⟨ 𝑋 , ( 𝐹𝑋 ) ⟩ } ) = 𝐹 )

Proof

Step Hyp Ref Expression
1 resdmdfsn ( 𝐹 ↾ ( V ∖ { 𝑋 } ) ) = ( 𝐹 ↾ ( dom 𝐹 ∖ { 𝑋 } ) )
2 1 a1i ( ( Fun 𝐹𝑋 ∈ dom 𝐹 ) → ( 𝐹 ↾ ( V ∖ { 𝑋 } ) ) = ( 𝐹 ↾ ( dom 𝐹 ∖ { 𝑋 } ) ) )
3 2 uneq1d ( ( Fun 𝐹𝑋 ∈ dom 𝐹 ) → ( ( 𝐹 ↾ ( V ∖ { 𝑋 } ) ) ∪ { ⟨ 𝑋 , ( 𝐹𝑋 ) ⟩ } ) = ( ( 𝐹 ↾ ( dom 𝐹 ∖ { 𝑋 } ) ) ∪ { ⟨ 𝑋 , ( 𝐹𝑋 ) ⟩ } ) )
4 funfn ( Fun 𝐹𝐹 Fn dom 𝐹 )
5 fnsnsplit ( ( 𝐹 Fn dom 𝐹𝑋 ∈ dom 𝐹 ) → 𝐹 = ( ( 𝐹 ↾ ( dom 𝐹 ∖ { 𝑋 } ) ) ∪ { ⟨ 𝑋 , ( 𝐹𝑋 ) ⟩ } ) )
6 4 5 sylanb ( ( Fun 𝐹𝑋 ∈ dom 𝐹 ) → 𝐹 = ( ( 𝐹 ↾ ( dom 𝐹 ∖ { 𝑋 } ) ) ∪ { ⟨ 𝑋 , ( 𝐹𝑋 ) ⟩ } ) )
7 3 6 eqtr4d ( ( Fun 𝐹𝑋 ∈ dom 𝐹 ) → ( ( 𝐹 ↾ ( V ∖ { 𝑋 } ) ) ∪ { ⟨ 𝑋 , ( 𝐹𝑋 ) ⟩ } ) = 𝐹 )