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 o. 2nd ) )
10 1 9 eqtri
 |-  F = frecs ( R , A , ( G o. 2nd ) )
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 )