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I am trying to prove A property of a cyclic order on a ring.

In order to do it, I need the last two properties (lemmas 1.11 and 1.12) in this question. I separated them from the original theorem because they are not related to rings, and applicable to any cyclically ordered group.

Please help me to find if the statements are correct.
Some of the intermediate lemmas are also used in the Quadrants of a cyclically ordered group.

If lemmas 1.11 and 1.12 are correct, then there is a question about the "magic" number $4$:
it looks like any circle with a positive element can always be split onto not more than $4$ positive segments (The rule of three steps for a cyclically ordered group);
at the same time, the logic of lemmas 1.11 and 1.12 works only up to $4$ positive elements.
Is this just a coincidence?

I am using the Cycle notation for cyclic orders wherever it is convenient.

Definition 1.1. An element $x$ of a cyclically ordered group is positive iff $[0, x, -x]$.

Definition 1.2. An element $x$ of a cyclically ordered group is negative iff $[0, -x, x]$.

Lemma 1.1. If an element $a$ of a cyclically ordered group is positive (negative), then element $-a$ is negative (positive).

Proof: follows from the definitions of positive and negative elements.

Lemma 1.2. If $a$ and $b$ are positive elements of a cyclically ordered group, then $a + b \ne 0$ and $b + a \ne 0$.

Proof:

  • Assuming $a + b = 0$ or $b + a = 0$: $a = -b$; $b = -a$;
  • Lemma 1.1: $a$ and $b$ are negative, contradiction.

Corollary 1.2. If element $a$ of a cyclically ordered group is positive, then $2a \ne 0$.

Lemma 1.3. If $a$ and $b$ are positive elements of a cyclically ordered group, and $a + a = b + b$, then $a = b$.

Proof:

  • Let $a + a = b + b = c$, $[0, a, b]$;
  • Compatibility with $a$: $[0, a, b] \implies [a, c, a + b]$;
  • Compatibility with $b$: $[0, a, b] \implies [b, a + b, c]$;
  • 4-Cycle: $[a, c, a + b] \land [b, a + b, c] \iff [a, c, b, a + b] \implies [a, c, b]$;
  • 4-Cycle: $[0, a, b] \land [a, c, b] \iff [0, a, c, b] \implies [b, 0, c]$;
  • Compatibility with $−b$: $[b, 0, c] \implies [0, -b, b]$;
  • $b$ is negative, contradiction.

Lemma 1.4. If $a$ is a positive element of a cyclically ordered group, and $[0, a', a]$ for an element $a'$, then $a'$ is positive.

Proof:

  • Negation: $[0, a', a] \iff [0, -a ,-a']$;
  • Transitivity: $[0, a, -a] \land [0, -a, -a'] \implies [0, a, -a']$;
  • Transitivity: $[0, a', a] \land [0, a, -a'] \implies [0, a', -a']$.

Lemma 1.5. If $a$ and $b$ are positive elements of a cyclically ordered group such that $[0, a, b]$, and $[a, c, b]$ for an element $c$, then $c$ is positive.

Proof:

  • 4-Cycle: $[0, a, b] \land [a, c, b] \iff [0, a, c, b] \implies [0, c, b]$;
  • Lemma 1.4: $[0, c, b] \implies$ $c$ is positive.

Lemma 1.6. If $a$ and $b$ are positive elements of a cyclically ordered group, then $[0, a, a + b]$, $[0, a, b + a]$, $[0, b, a + b]$, $[0, b, b + a]$.

Proof:

  • Lemma 1.2: $a + b \ne 0$, $b + a \ne 0$;
  • Assuming $[0, a + b, a]$ or $[0, b + a, a]$:
    • Compatibility with $–a$: $[-a, b, 0]$;
    • Cyclicity: $[-a, b, 0] \iff [0, -a, b]$;
    • Lemma 1.4: $[0, -a, b] \implies -a$ is positive;
    • Lemma 1.1: $a$ is negative, contradiction.
  • Assuming $[0, a + b, b]$ or $[0, b + a, b]$:
    • Compatibility with $–b$: $[-b, a, 0]$;
    • Cyclicity: $[-b, a, 0] \iff [0, -b, a]$;
    • Lemma 1.4: $[0, -b, a] \implies -b$ is positive;
    • Lemma 1.1: $b$ is negative, contradiction.

Corollary 1.6. If $a$ is a positive element of a cyclically ordered group, then $[0, a, 2a]$.

Lemma 1.7. If $a$ is a positive element of a cyclically ordered group, and $[0,b,a]$ for an element $b$, then $(a - b)$ is positive, and $[0, a - b, a]$, where $a - b \equiv a + (-b)$.

Proof:

  • Compatibility with $-b$: $[0, b, a] \implies [-b, 0, a - b]$;
  • Lemma 1.4: $[0, b, a] \implies b$ is positive;
  • Lemma 1.6: $[0, a, a + b]$;
  • Compatibility with $-b$: $[0, a, a + b] \implies [-b, a - b, a]$;
  • 4-Cycle: $[−b, 0, a − b] \land [−b, a − b, a] \iff [−b , 0, a − b, a] \implies [0, a − b, a]$;
  • Lemma 1.4: $[0, a - b, a] \implies (a - b)$ is positive.

Lemma 1.8. If $a$ and $b$ are positive elements of a cyclically ordered group, and $[0, a', a]$ for an element $a'$, then $[0, a' + b, a + b]$ and $[0, b + a', b + a]$.

Proof:

  • Compatibility with $b$: $[0, a', a] \implies [b, a' + b, a + b]$ (or $[b, b + a', b + a]$);
  • Lemma 1.6: $[0, b, a + b]$ (or $[0, b, b + a]$);
  • 4-Cycle: $[b, a′ + b, a + b] \land [0, b, a + b] \iff [0, b, a′ + b, a + b] \implies [0, a′ + b, a + b]$;
    (or $[b, b + a′, b + a] \land [0, b, b + a] \iff [0, b, b + a′, b + a] \implies [0, b + a′, b + a]$).

Corollary 1.8. If $b$ is a positive element of a cyclically ordered group, and $[0, a, b]$ for an element $a$, then $[0, 2a, a + b]$, $[0, 2a, b + a]$, $[0, a + b, 2b]$, and $[0, b + a, 2b]$.

Lemma 1.9. If $a$ and $b$ are positive elements of a cyclically ordered group, and $[0, a', a]$ and $[0, b', b]$ for some elements $a'$ and $b'$, then $[0, a' + b', a + b]$ and $[0, b' + a', b + a]$.

Proof:

  • Lemma 1.4: $[0, b', b] \implies b'$ is positive;
  • Lemma 1.8: $[0, a', a] \implies [0, a' + b', a + b']$ (or $[0, b' + a', b' + a]$);
  • Lemma 1.8: $[0, b', b] \implies [0, a + b', a + b]$ (or $[0, b' + a, b + a]$);
  • Transitivity: $[0, a' + b', a + b'] \land [0, a + b', a + b] \implies [0, a' + b', a + b]$
    (or $[0, b' + a', b' + a] \land [0, b' + a, b + a] \implies [0, b' + a', b + a]$).

Corollary 1.9. If $b$ is a positive element of a cyclically ordered group, and $[0, a, b]$ for an element $a$, then $[0, 2a, 2b]$.

Lemma 1.10. If $[0, a', a]$, $[0, b', b]$, and $a + b = 0$ for some elements $a$, $a'$, $b$, $b'$ of a cyclically ordered group, then $a' + b' \ne 0$.

Proof:

  1. $a + b = 0 \implies b = -a$;
  2. Assuming $a' + b' = 0$; $b' = -a'$;
    • Applying 1 and 2 to $[0, b', b]$: $[0, -a', -a]$;
    • Negation: $[0, a', a] \iff [0, -a, -a']$, contradiction.

Lemma 1.11. If $a$, $b$, and $c$ are positive elements of a cyclically ordered group such that $a + b + c = 0$, and $[0, a', a]$, $[0, b', b]$, $[0, c', c]$ for some elements $a'$, $b'$, $c'$, then $a' + b' + c' \ne 0$.

Proof:

  • Lemma 1.9: $[0, a', a] \land [0, b', b] \implies [0, a' + b', a + b]$;
  • Lemma 1.10: $[0, a' + b', a + b] \land [0, c', c] \land a + b + c = 0 \implies a' + b' + c' \ne 0$.

Lemma 1.12. If $a$, $b$, $c$, $d$ are positive elements of a cyclically ordered group such that $a + b + c + d = 0$, and $[0, a', a]$, $[0, b', b]$, $[0, c', c]$, $[0, d', d]$ for some elements $a'$, $b'$, $c'$, $d'$, then $a' + b' + c' + d' \ne 0$.

Proof:

  • Lemma 1.9: $[0, a', a] \land [0, b', b] \implies [0, a' + b', a + b]$;
  • Lemma 1.9: $[0, c', c] \land [0, d', d] \implies [0, c' + d', c + d]$;
  • Lemma 1.10: $[0, a' + b', a + b] \land [0, c' + d', c + d] \land a + b + c + d = 0 \implies a' + b' + c' + d' \ne 0$.

Note: The logic cannot be extended onto five or more positive elements: in $\mathbb Z_{10}$ element $4$ is positive, $4 + 4 + 4 + 4 + 4 = 0$, $[0,2,4]$, but $2 + 2 + 2 + 2 + 2 = 0$.

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  • $\begingroup$ What is a cyclically ordered group/ring? $\endgroup$ – darij grinberg May 23 at 14:33
  • $\begingroup$ @darijgrinberg: en.wikipedia.org/wiki/Cyclically_ordered_group. There is no common accepted definition for cyclically ordered rings, but we can talk about rings with cyclically ordered additive group. $\endgroup$ – Alex C May 23 at 14:40
  • $\begingroup$ I removed term "cyclically ordered ring". $\endgroup$ – Alex C May 23 at 15:35

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