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|xOnfFnxyxfy=Ffy
tfrlem.3 C=recsFdomrecsFFrecsF
Assertion tfrlem10 domrecsFOnCFnsucdomrecsF

Proof

Step Hyp Ref Expression
1 tfrlem.1 A=f|xOnfFnxyxfy=Ffy
2 tfrlem.3 C=recsFdomrecsFFrecsF
3 fvex FrecsFV
4 funsng domrecsFOnFrecsFVFundomrecsFFrecsF
5 3 4 mpan2 domrecsFOnFundomrecsFFrecsF
6 1 tfrlem7 FunrecsF
7 5 6 jctil domrecsFOnFunrecsFFundomrecsFFrecsF
8 3 dmsnop domdomrecsFFrecsF=domrecsF
9 8 ineq2i domrecsFdomdomrecsFFrecsF=domrecsFdomrecsF
10 1 tfrlem8 OrddomrecsF
11 orddisj OrddomrecsFdomrecsFdomrecsF=
12 10 11 ax-mp domrecsFdomrecsF=
13 9 12 eqtri domrecsFdomdomrecsFFrecsF=
14 funun FunrecsFFundomrecsFFrecsFdomrecsFdomdomrecsFFrecsF=FunrecsFdomrecsFFrecsF
15 7 13 14 sylancl domrecsFOnFunrecsFdomrecsFFrecsF
16 8 uneq2i domrecsFdomdomrecsFFrecsF=domrecsFdomrecsF
17 dmun domrecsFdomrecsFFrecsF=domrecsFdomdomrecsFFrecsF
18 df-suc sucdomrecsF=domrecsFdomrecsF
19 16 17 18 3eqtr4i domrecsFdomrecsFFrecsF=sucdomrecsF
20 df-fn recsFdomrecsFFrecsFFnsucdomrecsFFunrecsFdomrecsFFrecsFdomrecsFdomrecsFFrecsF=sucdomrecsF
21 15 19 20 sylanblrc domrecsFOnrecsFdomrecsFFrecsFFnsucdomrecsF
22 2 fneq1i CFnsucdomrecsFrecsFdomrecsFFrecsFFnsucdomrecsF
23 21 22 sylibr domrecsFOnCFnsucdomrecsF