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 ↾ Pred t++ R ↾ A A z ∪ z z G F ↾ Pred R A z = F ↾ Pred t++ R ↾ A A z ∪ z z G F ↾ Pred R A z
7 simpl ⊢ R Fr A ∧ R Se A → R Fr A
8 predres ⊢ Pred R A z = Pred R ↾ A A z
9 relres ⊢ Rel ⁡ R ↾ A
10 ssttrcl ⊢ Rel ⁡ R ↾ A → R ↾ A ⊆ t++ R ↾ A
11 predrelss ⊢ R ↾ A ⊆ t++ R ↾ A → Pred R ↾ A A z ⊆ Pred t++ R ↾ A A z
12 9 10 11 mp2b ⊢ Pred R ↾ A A z ⊆ Pred t++ R ↾ A A z
13 8 12 eqsstri ⊢ Pred R A z ⊆ Pred t++ R ↾ A A z
14 13 a1i ⊢ R Fr A ∧ R Se A ∧ z ∈ A → Pred R A z ⊆ Pred t++ R ↾ A A z
15 frrlem16 ⊢ R Fr A ∧ R Se A ∧ z ∈ A → ∀ a ∈ Pred t++ R ↾ A A z Pred R A a ⊆ Pred t++ R ↾ A A z
16 ttrclse ⊢ R Se A → t++ R ↾ A Se A
17 setlikespec ⊢ z ∈ A ∧ t++ R ↾ A Se A → Pred t++ R ↾ A A z ∈ V
18 17 ancoms ⊢ t++ R ↾ A Se A ∧ z ∈ A → Pred t++ R ↾ A A z ∈ V
19 16 18 sylan ⊢ R Se A ∧ z ∈ A → Pred t++ R ↾ A A z ∈ V
20 19 adantll ⊢ R Fr A ∧ R Se A ∧ z ∈ A → Pred t++ R ↾ A A z ∈ V
21 predss ⊢ Pred t++ R ↾ A A z ⊆ A
22 21 a1i ⊢ R Fr A ∧ R Se A ∧ z ∈ A → Pred t++ R ↾ A A z ⊆ A
23 difss ⊢ A ∖ dom ⁡ F ⊆ A
24 frmin ⊢ R Fr A ∧ R Se A ∧ A ∖ dom ⁡ F ⊆ A ∧ A ∖ dom ⁡ F ≠ ∅ → ∃ z ∈ A ∖ dom ⁡ F Pred R A ∖ dom ⁡ F z = ∅
25 23 24 mpanr1 ⊢ R Fr A ∧ R Se A ∧ A ∖ dom ⁡ F ≠ ∅ → ∃ z ∈ A ∖ dom ⁡ F Pred R A ∖ dom ⁡ F z = ∅
26 3 1 4 6 7 14 15 20 22 25 frrlem14 ⊢ R Fr A ∧ R Se A → dom ⁡ F = A
27 df-fn ⊢ F Fn A ↔ Fun ⁡ F ∧ dom ⁡ F = A
28 5 26 27 sylanbrc ⊢ R Fr A ∧ R Se A → F Fn A