Metamath Proof Explorer


Theorem bnj1522

Description: Well-founded recursion, part 3 of 3. The proof has been taken from Chapter 4 of Don Monk's notes on Set Theory. See http://euclid.colorado.edu/~monkd/setth.pdf . (Contributed by Jonathan Ben-Naim, 3-Jun-2011) (New usage is discouraged.)

Ref Expression
Hypotheses bnj1522.1 ⊢ B = d | d ⊆ A ∧ ∀ x ∈ d pred x A R ⊆ d
bnj1522.2 ⊢ Y = x f ↾ pred x A R
bnj1522.3 ⊢ C = f | ∃ d ∈ B f Fn d ∧ ∀ x ∈ d f ⁡ x = G ⁡ Y
bnj1522.4 ⊢ F = ⋃ C
Assertion bnj1522 ⊢ R FrSe A ∧ H Fn A ∧ ∀ x ∈ A H ⁡ x = G ⁡ x H ↾ pred x A R → F = H

Proof

Step Hyp Ref Expression
1 bnj1522.1 ⊢ B = d | d ⊆ A ∧ ∀ x ∈ d pred x A R ⊆ d
2 bnj1522.2 ⊢ Y = x f ↾ pred x A R
3 bnj1522.3 ⊢ C = f | ∃ d ∈ B f Fn d ∧ ∀ x ∈ d f ⁡ x = G ⁡ Y
4 bnj1522.4 ⊢ F = ⋃ C
5 biid ⊢ R FrSe A ∧ H Fn A ∧ ∀ x ∈ A H ⁡ x = G ⁡ x H ↾ pred x A R ↔ R FrSe A ∧ H Fn A ∧ ∀ x ∈ A H ⁡ x = G ⁡ x H ↾ pred x A R
6 biid ⊢ R FrSe A ∧ H Fn A ∧ ∀ x ∈ A H ⁡ x = G ⁡ x H ↾ pred x A R ∧ F ≠ H ↔ R FrSe A ∧ H Fn A ∧ ∀ x ∈ A H ⁡ x = G ⁡ x H ↾ pred x A R ∧ F ≠ H
7 biid ⊢ R FrSe A ∧ H Fn A ∧ ∀ x ∈ A H ⁡ x = G ⁡ x H ↾ pred x A R ∧ F ≠ H ∧ x ∈ A ∧ F ⁡ x ≠ H ⁡ x ↔ R FrSe A ∧ H Fn A ∧ ∀ x ∈ A H ⁡ x = G ⁡ x H ↾ pred x A R ∧ F ≠ H ∧ x ∈ A ∧ F ⁡ x ≠ H ⁡ x
8 eqid ⊢ x ∈ A | F ⁡ x ≠ H ⁡ x = x ∈ A | F ⁡ x ≠ H ⁡ x
9 biid ⊢ R FrSe A ∧ H Fn A ∧ ∀ x ∈ A H ⁡ x = G ⁡ x H ↾ pred x A R ∧ F ≠ H ∧ x ∈ A ∧ F ⁡ x ≠ H ⁡ x ∧ y ∈ x ∈ A | F ⁡ x ≠ H ⁡ x ∧ ∀ z ∈ x ∈ A | F ⁡ x ≠ H ⁡ x ¬ z R y ↔ R FrSe A ∧ H Fn A ∧ ∀ x ∈ A H ⁡ x = G ⁡ x H ↾ pred x A R ∧ F ≠ H ∧ x ∈ A ∧ F ⁡ x ≠ H ⁡ x ∧ y ∈ x ∈ A | F ⁡ x ≠ H ⁡ x ∧ ∀ z ∈ x ∈ A | F ⁡ x ≠ H ⁡ x ¬ z R y
10 1 2 3 4 5 6 7 8 9 bnj1523 ⊢ R FrSe A ∧ H Fn A ∧ ∀ x ∈ A H ⁡ x = G ⁡ x H ↾ pred x A R → F = H