Theorem - The contact number is sorted horizontally only be a natural straight line that gives a natural straight line

proof -\(\displaystyle 1\rightarrow 1\)

4\(\displaystyle {+_1^{\underline0}}\)2=2

4\(\displaystyle {+_1^{\underline1}}\)2=(1,1)

4\(\displaystyle {+_1^{\underline2}}\)2=2

4\(\displaystyle {+_1^{\underline3}}\)2=(3,1)

4\(\displaystyle {+_1^{\underline4}}\)2=6 or 4+2=6

\(\displaystyle +_1\) - addition rule 1

(CM.) - There are no "addition rule 1" only when the contact point number, Axiom

we were the first form of addition, the advantage of my mathematics

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question for trained mathematicians, which is a procedure to be applied in order to get (a sign?)

4 ? 2=2

4 ? 2=(1,1)

4 ? 2 =2

4 ? 2=(3,1)

4+2=6 This is known

more complex

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (1,1,1,2) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline0}}3 \underline25\)=(1,1,1,2)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (1,4) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline1}}3 \underline25\)=(1,4)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (2,1,1,2,1) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline2}}3 \underline25\)=(2,1,1,2,1)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (3,4,2) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline3}}3 \underline25\) =(3,4,2)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (6,4,1) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline4}}3 \underline25\)=(6,4,1)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (4,1,1,1,2) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline5}}3 \underline25\)=(4,1,1,1,2)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (4,3) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline6}}3 \underline25\)=(4,3)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (4,1,1,4) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline7}}3 \underline25\)=(4,1,1,4)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (4,2,2,5) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline8}}3 \underline25\)=(4,2,2,5)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (4,5,5) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline9}}3 \underline25\)=(4,5,5)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (4,3,1,5) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline{10}}}3 \underline25\)=(4,3,1,5)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (4,3,1,5) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline{11}}}3 \underline25\)=(4,3,1,5)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (4,3,2,5) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline{12}}}3 \underline25\)=(4,3,2,5)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (4,3,3,5) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline{13}}}3 \underline25\)=(4,3,3,5)

{0,4}\(\displaystyle \cup\){6,9}\(\displaystyle \cup\){11,13} ? {0,3}\(\displaystyle \cup\){5,10} = (4,3,3,5) my solution and notation \(\displaystyle 4\underline23\underline22{+_1^{\underline{14}}}3 \underline25\)=(4,3,3,5)