3.how to solve this current knowledge of mathematics:

along a (20m) ,deleted between 10 m and 15 m (b=5m) , wet get c

20m-(.5m)=¤10m(5m)5m¤
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Presupposition-In the exp

ression a

- (.b)c=d , d+(s.0))1

_{1} or d+(s.0)number (more) from 1

_{1}
and d+(.z)1

_{1}¤¤e or d+(.z) number ( more ) from 1

_{1}¤¤e , e={(f),(f)(f),(f)(f)(f),...}

Process:

P1 5

- (.0)5=5+(s.0)1

_{1} , 5

- (.0)5<6

_{1}
¤2(2)2¤

- (.1)¤1(1)2¤=¤1(2)1¤+(.z)1

_{1}¤¤(2)

¤2(2)2¤

- (.1)¤1(1)2¤<2¤¤(2)+(.z)1

_{1}¤¤(2) , ¤2(2)2¤ - (.1)¤1/1)2¤<3

_{1}¤¤(2)

P2 5 - (.0)5=5+(s.0)4 , 5 - (.0)5<9

¤2(2)2¤

- (.1)¤1(1)2¤=¤1(2)1¤+(.z)7¤¤(2)

¤2(2)2¤

- (.1)¤1(1)2)¤<2¤¤(2)+(.z)7¤¤(2), ¤2(2)2¤ - (.1)¤1(1)2¤<9¤¤(2)

...

General form a

- (.b)c=d-(s.0)1

_{1} ,a

- (s.b)c<g

_{1}
a

- (.b)c=d+(s.0)g , a

- (.b)c<l

a

- (.b)c=d+(s.0)k

_{p}g , a

- (.b)c<h

_{i}j ...

a

-(.b)c=d+(.z)1

_{1}¤¤e , a

- (.b)c<s¤¤e+(.z)1

_{1}¤¤e , a

- (s.b)c<g

_{1}¤¤e

a

- (.b)c=d+(.z)g¤¤e , a

- (.b)c<s¤¤¤e+(.z)g¤¤e , a

- (.b)c<l¤¤e

a

- (.b)c=d+(.z)k

_{p}g¤¤e , a

- (.b)c<s¤¤e+(.z)k

_{p}g¤¤e , a

- (.b)c<h

_{i}j¤¤e ...

[S29]-left inequality opposite subtraction

CM-[S29]-does no know

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Presupposition-In the exp

ression a

- (.b)c=d , d-(s.0/s.0))1

_{1}p or d+(s.0/s.0)number (more) from

1

_{1}p and d-(.z)1

_{1}p¤¤e or d-(.z) number ( more ) from 1

_{1}¤¤e , e={(f),(f)(f),(f)(f)(f),...}

Process:

P1 5

- (.0)5=5 -(s.0/s.0))1

_{1}5 , 5

- (.0)5>0

_{1}4

¤2(2)2¤

- (.1)¤1(1)2¤=¤1(2)1¤-(.z)2

_{1}1¤¤(2)

¤2(2)2¤

- (.1)¤1(1)2¤>2¤¤(2)-(.z)2

_{1}1¤¤(2) , ¤2(2)2¤

- (.1)¤1(1)2¤>0

_{1}1¤¤(2)

P2 5

- (.0)5=5-(s.0/s.0))1 , 5

- (.0)5>1

¤2(2)2¤

- (.1)¤1(1)2¤=¤1(2)1¤-(.z)1¤¤(2)

¤2(2)2¤

- (.1)¤1(1)2¤>2¤¤(2)-(.z)1¤¤(2) , ¤2(2)2¤

- (.1)¤1(1)2¤>1¤¤(2)

...

General form a

- (.b)c=d-(s.0/s.0)1

_{1}p ,a

- (s.b)c>g

_{1}l

a

- (.b)c=d-(s.0/s.0))g , a

- (.b)c>l

a

- (.b)c=d-(s.0/s.0))k

_{s}g , a

- (.b)c>h

_{i}j ...

a

- (.b)c=d-(.z)1

_{1}p¤¤e , a

- (.b)c>s¤¤e-(.z)1

_{1}p¤¤e , a

- (s.b)c>g

_{1}k¤¤e

a

- (.b)c=d-(.z)g¤¤e , a

- (.b)c>s¤¤e-(.z)g¤¤e , a

- (.b)c>l¤¤e

a

- (.b)c=d-(.z)k

_{p}g¤¤e , a

- (.b)c>s¤¤e-(.z)k

_{p}g¤¤e , a

- (.b)c>h

_{i}j¤¤e ...

[S30]-right inequality opposite subtraction

CM-[S30]-does no know

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