Metamath Proof Explorer


Theorem wfrlem16OLD

Description: Lemma for well-ordered recursion. If z is R minimal in ( A \ dom F ) , then C is acceptable and thus a subset of F , but dom C is bigger than dom F . Thus, z cannot be minimal, so ( A \ dom F ) must be empty, and (due to wfrdmssOLD ), dom F = A . Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011)

Ref Expression
Hypotheses wfrlem13OLD.1 RWeA
wfrlem13OLD.2 RSeA
wfrlem13OLD.3 F=wrecsRAG
wfrlem13OLD.4 C=FzGFPredRAz
Assertion wfrlem16OLD domF=A

Proof

Step Hyp Ref Expression
1 wfrlem13OLD.1 RWeA
2 wfrlem13OLD.2 RSeA
3 wfrlem13OLD.3 F=wrecsRAG
4 wfrlem13OLD.4 C=FzGFPredRAz
5 3 wfrdmssOLD domFA
6 eldifn zAdomF¬zdomF
7 ssun2 zdomFz
8 vsnid zz
9 7 8 sselii zdomFz
10 4 dmeqi domC=domFzGFPredRAz
11 dmun domFzGFPredRAz=domFdomzGFPredRAz
12 fvex GFPredRAzV
13 12 dmsnop domzGFPredRAz=z
14 13 uneq2i domFdomzGFPredRAz=domFz
15 10 11 14 3eqtri domC=domFz
16 9 15 eleqtrri zdomC
17 1 2 3 4 wfrlem15OLD zAdomFPredRAdomFz=Cf|xfFnxxAyxPredRAyxyxfy=GfPredRAy
18 elssuni Cf|xfFnxxAyxPredRAyxyxfy=GfPredRAyCf|xfFnxxAyxPredRAyxyxfy=GfPredRAy
19 17 18 syl zAdomFPredRAdomFz=Cf|xfFnxxAyxPredRAyxyxfy=GfPredRAy
20 dfwrecsOLD wrecsRAG=f|xfFnxxAyxPredRAyxyxfy=GfPredRAy
21 3 20 eqtri F=f|xfFnxxAyxPredRAyxyxfy=GfPredRAy
22 19 21 sseqtrrdi zAdomFPredRAdomFz=CF
23 dmss CFdomCdomF
24 22 23 syl zAdomFPredRAdomFz=domCdomF
25 24 sseld zAdomFPredRAdomFz=zdomCzdomF
26 16 25 mpi zAdomFPredRAdomFz=zdomF
27 6 26 mtand zAdomF¬PredRAdomFz=
28 27 nrex ¬zAdomFPredRAdomFz=
29 df-ne AdomF¬AdomF=
30 difss AdomFA
31 1 2 tz6.26i AdomFAAdomFzAdomFPredRAdomFz=
32 30 31 mpan AdomFzAdomFPredRAdomFz=
33 29 32 sylbir ¬AdomF=zAdomFPredRAdomFz=
34 28 33 mt3 AdomF=
35 ssdif0 AdomFAdomF=
36 34 35 mpbir AdomF
37 5 36 eqssi domF=A