Metamath Proof Explorer


Theorem setsfun

Description: A structure with replacement is a function if the original structure is a function. (Contributed by AV, 7-Jun-2021)

Ref Expression
Assertion setsfun GVFunGIUEWFunGsSetIE

Proof

Step Hyp Ref Expression
1 funres FunGFunGVdomIE
2 1 adantl GVFunGFunGVdomIE
3 2 adantr GVFunGIUEWFunGVdomIE
4 funsng IUEWFunIE
5 4 adantl GVFunGIUEWFunIE
6 dmres domGVdomIE=VdomIEdomG
7 6 ineq1i domGVdomIEdomIE=VdomIEdomGdomIE
8 in32 VdomIEdomGdomIE=VdomIEdomIEdomG
9 disjdifr VdomIEdomIE=
10 9 ineq1i VdomIEdomIEdomG=domG
11 0in domG=
12 8 10 11 3eqtri VdomIEdomGdomIE=
13 7 12 eqtri domGVdomIEdomIE=
14 13 a1i GVFunGIUEWdomGVdomIEdomIE=
15 funun FunGVdomIEFunIEdomGVdomIEdomIE=FunGVdomIEIE
16 3 5 14 15 syl21anc GVFunGIUEWFunGVdomIEIE
17 opex IEV
18 17 a1i FunGIEV
19 setsvalg GVIEVGsSetIE=GVdomIEIE
20 18 19 sylan2 GVFunGGsSetIE=GVdomIEIE
21 20 funeqd GVFunGFunGsSetIEFunGVdomIEIE
22 21 adantr GVFunGIUEWFunGsSetIEFunGVdomIEIE
23 16 22 mpbird GVFunGIUEWFunGsSetIE