Metamath Proof Explorer


Theorem fpwwe2lem9

Description: Lemma for fpwwe2 . Given two well-orders <. X , R >. and <. Y , S >. of parts of A , one is an initial segment of the other. (Contributed by Mario Carneiro, 15-May-2015) (Revised by AV, 20-Jul-2024)

Ref Expression
Hypotheses fpwwe2.1 W=xr|xArx×xrWexyx[˙r-1y/u]˙uFru×u=y
fpwwe2.2 φAV
fpwwe2.3 φxArx×xrWexxFrA
fpwwe2lem9.4 φXWR
fpwwe2lem9.6 φYWS
Assertion fpwwe2lem9 φXYR=SY×XYXS=RX×Y

Proof

Step Hyp Ref Expression
1 fpwwe2.1 W=xr|xArx×xrWexyx[˙r-1y/u]˙uFru×u=y
2 fpwwe2.2 φAV
3 fpwwe2.3 φxArx×xrWexxFrA
4 fpwwe2lem9.4 φXWR
5 fpwwe2lem9.6 φYWS
6 eqid OrdIsoRX=OrdIsoRX
7 6 oicl OrddomOrdIsoRX
8 eqid OrdIsoSY=OrdIsoSY
9 8 oicl OrddomOrdIsoSY
10 ordtri2or2 OrddomOrdIsoRXOrddomOrdIsoSYdomOrdIsoRXdomOrdIsoSYdomOrdIsoSYdomOrdIsoRX
11 7 9 10 mp2an domOrdIsoRXdomOrdIsoSYdomOrdIsoSYdomOrdIsoRX
12 2 adantr φdomOrdIsoRXdomOrdIsoSYAV
13 3 adantlr φdomOrdIsoRXdomOrdIsoSYxArx×xrWexxFrA
14 4 adantr φdomOrdIsoRXdomOrdIsoSYXWR
15 5 adantr φdomOrdIsoRXdomOrdIsoSYYWS
16 simpr φdomOrdIsoRXdomOrdIsoSYdomOrdIsoRXdomOrdIsoSY
17 1 12 13 14 15 6 8 16 fpwwe2lem8 φdomOrdIsoRXdomOrdIsoSYXYR=SY×X
18 17 ex φdomOrdIsoRXdomOrdIsoSYXYR=SY×X
19 2 adantr φdomOrdIsoSYdomOrdIsoRXAV
20 3 adantlr φdomOrdIsoSYdomOrdIsoRXxArx×xrWexxFrA
21 5 adantr φdomOrdIsoSYdomOrdIsoRXYWS
22 4 adantr φdomOrdIsoSYdomOrdIsoRXXWR
23 simpr φdomOrdIsoSYdomOrdIsoRXdomOrdIsoSYdomOrdIsoRX
24 1 19 20 21 22 8 6 23 fpwwe2lem8 φdomOrdIsoSYdomOrdIsoRXYXS=RX×Y
25 24 ex φdomOrdIsoSYdomOrdIsoRXYXS=RX×Y
26 18 25 orim12d φdomOrdIsoRXdomOrdIsoSYdomOrdIsoSYdomOrdIsoRXXYR=SY×XYXS=RX×Y
27 11 26 mpi φXYR=SY×XYXS=RX×Y