Metamath Proof Explorer


Theorem frr1

Description: Law of general well-founded recursion, part one. This theorem and the following two drop the partial order requirement from fpr1 , fpr2 , and fpr3 , which requires using the axiom of infinity (Contributed by Scott Fenton, 11-Sep-2023)

Ref Expression
Hypothesis frr.1 F = frecs R A G
Assertion frr1 R Fr A R Se A F Fn A

Proof

Step Hyp Ref Expression
1 frr.1 F = frecs R A G
2 eqid a | b a Fn b b A c b Pred R A c b c b a c = c G a Pred R A c = a | b a Fn b b A c b Pred R A c b c b a c = c G a Pred R A c
3 2 frrlem1 a | b a Fn b b A c b Pred R A c b c b a c = c G a Pred R A c = f | x f Fn x x A y x Pred R A y x y x f y = y G f Pred R A y
4 3 1 frrlem15 R Fr A R Se A g a | b a Fn b b A c b Pred R A c b c b a c = c G a Pred R A c h a | b a Fn b b A c b Pred R A c b c b a c = c G a Pred R A c x g u x h v u = v
5 3 1 4 frrlem9 R Fr A R Se A Fun F
6 eqid F TrPred R A z z z G F Pred R A z = F TrPred R A z z z G F Pred R A z
7 simpl R Fr A R Se A R Fr A
8 setlikespec z A R Se A Pred R A z V
9 8 ancoms R Se A z A Pred R A z V
10 9 adantll R Fr A R Se A z A Pred R A z V
11 trpredpred Pred R A z V Pred R A z TrPred R A z
12 10 11 syl R Fr A R Se A z A Pred R A z TrPred R A z
13 frrlem16 R Fr A R Se A z A a TrPred R A z Pred R A a TrPred R A z
14 trpredex TrPred R A z V
15 14 a1i R Fr A R Se A z A TrPred R A z V
16 trpredss Pred R A z V TrPred R A z A
17 10 16 syl R Fr A R Se A z A TrPred R A z A
18 difss A dom F A
19 frmin R Fr A R Se A A dom F A A dom F z A dom F Pred R A dom F z =
20 18 19 mpanr1 R Fr A R Se A A dom F z A dom F Pred R A dom F z =
21 3 1 4 6 7 12 13 15 17 20 frrlem14 R Fr A R Se A dom F = A
22 df-fn F Fn A Fun F dom F = A
23 5 21 22 sylanbrc R Fr A R Se A F Fn A