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