Metamath Proof Explorer


Theorem wfrfun

Description: The "function" generated by the well-ordered recursion generator is indeed a function. Avoids the axiom of replacement. (Contributed by Scott Fenton, 21-Apr-2011) (Revised by Mario Carneiro, 26-Jun-2015) (Revised by Scott Fenton, 17-Nov-2024)

Ref Expression
Hypothesis wfrfun.1 F = wrecs R A G
Assertion wfrfun R We A R Se A Fun F

Proof

Step Hyp Ref Expression
1 wfrfun.1 F = wrecs R A G
2 wefr R We A R Fr A
3 2 adantr R We A R Se A R Fr A
4 weso R We A R Or A
5 sopo R Or A R Po A
6 4 5 syl R We A R Po A
7 6 adantr R We A R Se A R Po A
8 simpr R We A R Se A R Se A
9 df-wrecs wrecs R A G = frecs R A G 2 nd
10 1 9 eqtri F = frecs R A G 2 nd
11 10 fprfung R Fr A R Po A R Se A Fun F
12 3 7 8 11 syl3anc R We A R Se A Fun F