Metamath Proof Explorer


Theorem unblem2

Description: Lemma for unbnn . The value of the function F belongs to the unbounded set of natural numbers A . (Contributed by NM, 3-Dec-2003)

Ref Expression
Hypothesis unblem.2 F=recxVAsucxAω
Assertion unblem2 AωwωvAwvzωFzA

Proof

Step Hyp Ref Expression
1 unblem.2 F=recxVAsucxAω
2 fveq2 z=Fz=F
3 2 eleq1d z=FzAFA
4 fveq2 z=uFz=Fu
5 4 eleq1d z=uFzAFuA
6 fveq2 z=sucuFz=Fsucu
7 6 eleq1d z=sucuFzAFsucuA
8 omsson ωOn
9 sstr AωωOnAOn
10 8 9 mpan2 AωAOn
11 peano1 ω
12 eleq1 w=wvv
13 12 rexbidv w=vAwvvAv
14 13 rspcv ωwωvAwvvAv
15 11 14 ax-mp wωvAwvvAv
16 df-rex vAvvvAv
17 15 16 sylib wωvAwvvvAv
18 exsimpl vvAvvvA
19 17 18 syl wωvAwvvvA
20 n0 AvvA
21 19 20 sylibr wωvAwvA
22 onint AOnAAA
23 10 21 22 syl2an AωwωvAwvAA
24 1 fveq1i F=recxVAsucxAω
25 fr0g AArecxVAsucxAω=A
26 24 25 eqtr2id AAA=F
27 26 eleq1d AAAAFA
28 27 ibi AAFA
29 23 28 syl AωwωvAwvFA
30 unblem1 AωwωvAwvFuAAsucFuA
31 suceq y=xsucy=sucx
32 31 difeq2d y=xAsucy=Asucx
33 32 inteqd y=xAsucy=Asucx
34 suceq y=Fusucy=sucFu
35 34 difeq2d y=FuAsucy=AsucFu
36 35 inteqd y=FuAsucy=AsucFu
37 1 33 36 frsucmpt2 uωAsucFuAFsucu=AsucFu
38 37 eqcomd uωAsucFuAAsucFu=Fsucu
39 38 eleq1d uωAsucFuAAsucFuAFsucuA
40 39 ex uωAsucFuAAsucFuAFsucuA
41 40 ibd uωAsucFuAFsucuA
42 30 41 syl5 uωAωwωvAwvFuAFsucuA
43 42 expd uωAωwωvAwvFuAFsucuA
44 3 5 7 29 43 finds2 zωAωwωvAwvFzA
45 44 com12 AωwωvAwvzωFzA