differential propositional calculus : appendix 4


0.1 Detail of Calculation for the Difference Map

Detail of Calculation for Df=Ef+f
Ef|dxdy+f|dxdy=Df|dxdy Ef|dx(dy)+f|dx(dy)=Df|dx(dy) Ef|(dx)dy+f|(dx)dy=Df|(dx)dy Ef|(dx)(dy)+f|(dx)(dy)=Df|(dx)(dy)
f0 0+0=0 0+0=0 0+0=0 0+0=0
f1 xydxdy+(x)(y)dxdy=((x,y))dxdy x(y)dx(dy)+(x)(y)dx(dy)=(y)dx(dy) (x)y(dx)dy+(x)(y)(dx)dy=(x)(dx)dy (x)(y)(dx)(dy)+(x)(y)(dx)(dy)=0(dx)(dy)
f2 x(y)dxdy+(x)ydxdy=(x,y)dxdy xydx(dy)+(x)ydx(dy)=ydx(dy) (x)(y)(dx)dy+(x)y(dx)dy=(x)(dx)dy (x)y(dx)(dy)+(x)y(dx)(dy)=0(dx)(dy)
f4 (x)ydxdy+x(y)dxdy=(x,y)dxdy (x)(y)dx(dy)+x(y)dx(dy)=(y)dx(dy) xy(dx)dy+x(y)(dx)dy=x(dx)dy x(y)(dx)(dy)+x(y)(dx)(dy)=0(dx)(dy)
f8 (x)(y)dxdy+xydxdy=((x,y))dxdy (x)ydx(dy)+xydx(dy)=ydx(dy) x(y)(dx)dy+xy(dx)dy=x(dx)dy xy(dx)(dy)+xy(dx)(dy)=0(dx)(dy)
f3 xdxdy+(x)dxdy=1dxdy xdx(dy)+(x)dx(dy)=1dx(dy) (x)(dx)dy+(x)(dx)dy=0(dx)dy (x)(dx)(dy)+(x)(dx)(dy)=0(dx)(dy)
f12 (x)dxdy+xdxdy=1dxdy (x)dx(dy)+xdx(dy)=1dx(dy) x(dx)dy+x(dx)dy=0(dx)dy x(dx)(dy)+x(dx)(dy)=0(dx)(dy)
f6 (x,y)dxdy+(x,y)dxdy=0dxdy ((x,y))dx(dy)+(x,y)dx(dy)=1dx(dy) ((x,y))(dx)dy+(x,y)(dx)dy=1(dx)dy (x,y)(dx)(dy)+(x,y)(dx)(dy)=0(dx)(dy)
f9 ((x,y))dxdy+((x,y))dxdy=0dxdy (x,y)dx(dy)+((x,y))dx(dy)=1dx(dy) (x,y)(dx)dy+((x,y))(dx)dy=1(dx)dy ((x,y))(dx)(dy)+((x,y))(dx)(dy)=0(dx)(dy)
f5 ydxdy+(y)dxdy=1dxdy (y)dx(dy)+(y)dx(dy)=0dx(dy) y(dx)dy+(y)(dx)dy=1(dx)dy (y)(dx)(dy)+(y)(dx)(dy)=0(dx)(dy)
f10 (y)dxdy+ydxdy=1dxdy ydx(dy)+ydx(dy)=0dx(dy) (y)(dx)dy+y(dx)dy=1(dx)dy y(dx)(dy)+y(dx)(dy)=0(dx)(dy)
f7 ((x)(y))dxdy+(xy)dxdy=((x,y))dxdy ((x)y)dx(dy)+(xy)dx(dy)=ydx(dy) (x(y))(dx)dy+(xy)(dx)dy=x(dx)dy (xy)(dx)(dy)+(xy)(dx)(dy)=0(dx)(dy)
f11 ((x)y)dxdy+(x(y))dxdy=(x,y)dxdy ((x)(y))dx(dy)+(x(y))dx(dy)=(y)dx(dy) (xy)(dx)dy+(x(y))(dx)dy=x(dx)dy (x(y))(dx)(dy)+(x(y))(dx)(dy)=0(dx)(dy)
f13 (x(y))dxdy+((x)y)dxdy=(x,y)dxdy (xy)dx(dy)+((x)y)dx(dy)=ydx(dy) ((x)(y))(dx)dy+((x)y)(dx)dy=(x)(dx)dy ((x)y)(dx)(dy)+((x)y)(dx)(dy)=0(dx)(dy)
f14 (xy)dxdy+((x)(y))dxdy=((x,y))dxdy (x(y))dx(dy)+((x)(y))dx(dy)=(y)dx(dy) ((x)y)(dx)dy+((x)(y))(dx)dy=(x)(dx)dy ((x)(y))(dx)(dy)+((x)(y))(dx)(dy)=0(dx)(dy)
f15 1+1=0 1+1=0 1+1=0 1+1=0
Title differential propositional calculus : appendix 4
Canonical name DifferentialPropositionalCalculusAppendix4
Date of creation 2013-03-22 18:09:25
Last modified on 2013-03-22 18:09:25
Owner Jon Awbrey (15246)
Last modified by Jon Awbrey (15246)
Numerical id 7
Author Jon Awbrey (15246)
Entry type Application
Classification msc 53A40
Classification msc 39A12
Classification msc 34G99
Classification msc 03B44
Classification msc 03B05
Classification msc 03B42
Related topic DifferentialLogic
Related topic MinimalNegationOperator
Related topic PropositionalCalculus
Related topic ZerothOrderLogic