Metamath Proof Explorer


Theorem tfrlem10

Description: Lemma for transfinite recursion. We define class C by extending recs with one ordered pair. We will assume, falsely, that domain of recs is a member of, and thus not equal to, On . Using this assumption we will prove facts about C that will lead to a contradiction in tfrlem14 , thus showing the domain of recs does in fact equal On . Here we show (under the false assumption) that C is a function extending the domain of recs ( F ) by one. (Contributed by NM, 14-Aug-1994) (Revised by Mario Carneiro, 9-May-2015)

Ref Expression
Hypotheses tfrlem.1 ⊢ A = f | ∃ x ∈ On f Fn x ∧ ∀ y ∈ x f ⁡ y = F ⁡ f ↾ y
tfrlem.3 ⊢ C = recs ⁡ F ∪ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F
Assertion tfrlem10 ⊢ dom ⁡ recs ⁡ F ∈ On → C Fn suc ⁡ dom ⁡ recs ⁡ F

Proof

Step Hyp Ref Expression
1 tfrlem.1 ⊢ A = f | ∃ x ∈ On f Fn x ∧ ∀ y ∈ x f ⁡ y = F ⁡ f ↾ y
2 tfrlem.3 ⊢ C = recs ⁡ F ∪ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F
3 fvex ⊢ F ⁡ recs ⁡ F ∈ V
4 funsng ⊢ dom ⁡ recs ⁡ F ∈ On ∧ F ⁡ recs ⁡ F ∈ V → Fun ⁡ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F
5 3 4 mpan2 ⊢ dom ⁡ recs ⁡ F ∈ On → Fun ⁡ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F
6 1 tfrlem7 ⊢ Fun ⁡ recs ⁡ F
7 5 6 jctil ⊢ dom ⁡ recs ⁡ F ∈ On → Fun ⁡ recs ⁡ F ∧ Fun ⁡ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F
8 3 dmsnop ⊢ dom ⁡ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F = dom ⁡ recs ⁡ F
9 8 ineq2i ⊢ dom ⁡ recs ⁡ F ∩ dom ⁡ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F = dom ⁡ recs ⁡ F ∩ dom ⁡ recs ⁡ F
10 1 tfrlem8 ⊢ Ord ⁡ dom ⁡ recs ⁡ F
11 orddisj ⊢ Ord ⁡ dom ⁡ recs ⁡ F → dom ⁡ recs ⁡ F ∩ dom ⁡ recs ⁡ F = ∅
12 10 11 ax-mp ⊢ dom ⁡ recs ⁡ F ∩ dom ⁡ recs ⁡ F = ∅
13 9 12 eqtri ⊢ dom ⁡ recs ⁡ F ∩ dom ⁡ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F = ∅
14 funun ⊢ Fun ⁡ recs ⁡ F ∧ Fun ⁡ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F ∧ dom ⁡ recs ⁡ F ∩ dom ⁡ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F = ∅ → Fun ⁡ recs ⁡ F ∪ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F
15 7 13 14 sylancl ⊢ dom ⁡ recs ⁡ F ∈ On → Fun ⁡ recs ⁡ F ∪ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F
16 8 uneq2i ⊢ dom ⁡ recs ⁡ F ∪ dom ⁡ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F = dom ⁡ recs ⁡ F ∪ dom ⁡ recs ⁡ F
17 dmun ⊢ dom ⁡ recs ⁡ F ∪ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F = dom ⁡ recs ⁡ F ∪ dom ⁡ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F
18 df-suc ⊢ suc ⁡ dom ⁡ recs ⁡ F = dom ⁡ recs ⁡ F ∪ dom ⁡ recs ⁡ F
19 16 17 18 3eqtr4i ⊢ dom ⁡ recs ⁡ F ∪ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F = suc ⁡ dom ⁡ recs ⁡ F
20 df-fn ⊢ recs ⁡ F ∪ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F Fn suc ⁡ dom ⁡ recs ⁡ F ↔ Fun ⁡ recs ⁡ F ∪ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F ∧ dom ⁡ recs ⁡ F ∪ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F = suc ⁡ dom ⁡ recs ⁡ F
21 15 19 20 sylanblrc ⊢ dom ⁡ recs ⁡ F ∈ On → recs ⁡ F ∪ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F Fn suc ⁡ dom ⁡ recs ⁡ F
22 2 fneq1i ⊢ C Fn suc ⁡ dom ⁡ recs ⁡ F ↔ recs ⁡ F ∪ dom ⁡ recs ⁡ F F ⁡ recs ⁡ F Fn suc ⁡ dom ⁡ recs ⁡ F
23 21 22 sylibr ⊢ dom ⁡ recs ⁡ F ∈ On → C Fn suc ⁡ dom ⁡ recs ⁡ F