The premise leads to a contradiction. The only conclusion is that the premise is false.

proof.txt

Premise

- consider X, a non-zero integer with n digits;
- consider Y, a non-zero integer that is formed by moving the first digit from X to the position of the last digit and shifting all of the remaining digits one place to their immediate left;
- there is a value for X such that, when doubled, equals Y.

Axioms

- when you double an integer, you multiply each of its digits by 2;
- the product of each digit being multipled by 2 may be incremented by 1 if the digit to its immediate right was 5 or more.

Proof no digit other than the first digit can be odd and less than 5:

- after doubling the number, each digit must equal the digit that was to its immediate right (with the exception of the last digit);
- the only way for a digit to become odd when multipled by 2 is if it is incremented by 1;
- if the digit to its immediate right was less than 5, it won't be incremented;
- therefore, no digit other than the first digit can be odd and less than 5.

Proof that no digit other than the first digit can be even and more than 5;

- after doubling the number, each digit must equal the digit that was to its immediate right (with the exception of the last digit);
- every digit when multiplied by 2 is even;
- when a digit that is 5 or more is multipled by 2, the product of the digit to its immediate left being multipled by 2 is incremented by 1;
- a digit multipled by 2 and incremented by 1 must be odd;
- therefore, no digit other than the first digit can be even and more than 5.

Proof that the first digit is less than 5:

- if the first digit is 5 or more and multiplied by 2, this increases the total number of digits in the number by 1;
- the total number of digits in the number cannot increase;
- therefore, the first digit must be less than 5.

Proof that the first digit is even:

- a digit multipled by 2 is always even unless the digit to its immediate right doubles to more than 10;
- there is no digit to the immediate right of the last digit, so the last digit, when multipled by 2, will be even;
- the mod 10 of the product of the last digit being multipled by 2 must equal the first digit;
- therefore, the first digit must be even.

Proof that the first digit is 2 or 4:

- the first digit is even;
- the first digit is less than 5;
- therefore, the first digit must be 2 or 4.

Proof that the last digit is 2 or 7:

- the first digit must be 2 or 4;
- mod 10 of the product of the last digit multipled by 2 must be equal to the first digit;
- the numbers 1 and 6 when multiplied by 2 have a mod 10 of 2;
- the numbers 2 and 7 when multiplied by 2 have a mod 10 of 4;
- no digit can be odd and less than 5;
- no digit can be even and more than 5;
- therefore, the last digit must be 2 or 7.

Proof that the last digit cannot be 2:

- after doubling the number, each digit must equal the digit that was to its immediate right (with the exception of the last digit);
- the only way for the second last digit to be equal to 2 when multipled by 2 is if it is 1 or 6;
- no digit can be odd and less than 5;
- no digit can be even and more than 5;
- therefore, the last digit cannot be 2.

Proof that the last digit cannot be 7:

- after doubling the number, each digit must equal the digit that was to its immediate right (with the exception of the last digit);
- if the last digit is 7, the product of the second last digit when multipled by 2 will be incremented by 1;
- the product of the second last digit being multipled by 2 must be 6;
- the only way for the second last digit to be equal to 6 when multipled by 2 is if it is 3 or 8;
- no digit can be odd and less than 5;
- no digit can be even and more than 5;
- therefore, the last digit cannot be 7.