Metamath Proof Explorer


Theorem abvpropd

Description: If two structures have the same ring components, they have the same collection of absolute values. (Contributed by Mario Carneiro, 4-Oct-2015)

Ref Expression
Hypotheses abvpropd.1 φB=BaseK
abvpropd.2 φB=BaseL
abvpropd.3 φxByBx+Ky=x+Ly
abvpropd.4 φxByBxKy=xLy
Assertion abvpropd φAbsValK=AbsValL

Proof

Step Hyp Ref Expression
1 abvpropd.1 φB=BaseK
2 abvpropd.2 φB=BaseL
3 abvpropd.3 φxByBx+Ky=x+Ly
4 abvpropd.4 φxByBxKy=xLy
5 1 2 3 4 ringpropd φKRingLRing
6 1 2 eqtr3d φBaseK=BaseL
7 6 feq2d φf:BaseK0+∞f:BaseL0+∞
8 1 2 3 grpidpropd φ0K=0L
9 8 adantr φxB0K=0L
10 9 eqeq2d φxBx=0Kx=0L
11 10 bibi2d φxBfx=0x=0Kfx=0x=0L
12 4 fveqeq2d φxByBfxKy=fxfyfxLy=fxfy
13 3 fveq2d φxByBfx+Ky=fx+Ly
14 13 breq1d φxByBfx+Kyfx+fyfx+Lyfx+fy
15 12 14 anbi12d φxByBfxKy=fxfyfx+Kyfx+fyfxLy=fxfyfx+Lyfx+fy
16 15 anassrs φxByBfxKy=fxfyfx+Kyfx+fyfxLy=fxfyfx+Lyfx+fy
17 16 ralbidva φxByBfxKy=fxfyfx+Kyfx+fyyBfxLy=fxfyfx+Lyfx+fy
18 11 17 anbi12d φxBfx=0x=0KyBfxKy=fxfyfx+Kyfx+fyfx=0x=0LyBfxLy=fxfyfx+Lyfx+fy
19 18 ralbidva φxBfx=0x=0KyBfxKy=fxfyfx+Kyfx+fyxBfx=0x=0LyBfxLy=fxfyfx+Lyfx+fy
20 1 raleqdv φyBfxKy=fxfyfx+Kyfx+fyyBaseKfxKy=fxfyfx+Kyfx+fy
21 20 anbi2d φfx=0x=0KyBfxKy=fxfyfx+Kyfx+fyfx=0x=0KyBaseKfxKy=fxfyfx+Kyfx+fy
22 1 21 raleqbidv φxBfx=0x=0KyBfxKy=fxfyfx+Kyfx+fyxBaseKfx=0x=0KyBaseKfxKy=fxfyfx+Kyfx+fy
23 2 raleqdv φyBfxLy=fxfyfx+Lyfx+fyyBaseLfxLy=fxfyfx+Lyfx+fy
24 23 anbi2d φfx=0x=0LyBfxLy=fxfyfx+Lyfx+fyfx=0x=0LyBaseLfxLy=fxfyfx+Lyfx+fy
25 2 24 raleqbidv φxBfx=0x=0LyBfxLy=fxfyfx+Lyfx+fyxBaseLfx=0x=0LyBaseLfxLy=fxfyfx+Lyfx+fy
26 19 22 25 3bitr3d φxBaseKfx=0x=0KyBaseKfxKy=fxfyfx+Kyfx+fyxBaseLfx=0x=0LyBaseLfxLy=fxfyfx+Lyfx+fy
27 7 26 anbi12d φf:BaseK0+∞xBaseKfx=0x=0KyBaseKfxKy=fxfyfx+Kyfx+fyf:BaseL0+∞xBaseLfx=0x=0LyBaseLfxLy=fxfyfx+Lyfx+fy
28 5 27 anbi12d φKRingf:BaseK0+∞xBaseKfx=0x=0KyBaseKfxKy=fxfyfx+Kyfx+fyLRingf:BaseL0+∞xBaseLfx=0x=0LyBaseLfxLy=fxfyfx+Lyfx+fy
29 eqid AbsValK=AbsValK
30 29 abvrcl fAbsValKKRing
31 eqid BaseK=BaseK
32 eqid +K=+K
33 eqid K=K
34 eqid 0K=0K
35 29 31 32 33 34 isabv KRingfAbsValKf:BaseK0+∞xBaseKfx=0x=0KyBaseKfxKy=fxfyfx+Kyfx+fy
36 30 35 biadanii fAbsValKKRingf:BaseK0+∞xBaseKfx=0x=0KyBaseKfxKy=fxfyfx+Kyfx+fy
37 eqid AbsValL=AbsValL
38 37 abvrcl fAbsValLLRing
39 eqid BaseL=BaseL
40 eqid +L=+L
41 eqid L=L
42 eqid 0L=0L
43 37 39 40 41 42 isabv LRingfAbsValLf:BaseL0+∞xBaseLfx=0x=0LyBaseLfxLy=fxfyfx+Lyfx+fy
44 38 43 biadanii fAbsValLLRingf:BaseL0+∞xBaseLfx=0x=0LyBaseLfxLy=fxfyfx+Lyfx+fy
45 28 36 44 3bitr4g φfAbsValKfAbsValL
46 45 eqrdv φAbsValK=AbsValL