The **Lascar inequalities** say that if are finite tuples and is a set of parameters in a stable theory, then Here denotes the Lascar -rank of . On the left, denotes the usual ordinal sum, while on the right, denotes the **natural sum** of ordinals, defined by adding the coefficients in Cantor normal form:

There is also a related auxiliary result: if , then .

More generally, the same results hold in simple theories, with -rank rather than -rank. The same proofs work.

Recall the definition of Lascar rank: if and only if there is some such that is a forking extension of and .

We will use the following basic facts.

Fact 1: If , then , so Lascar rank is appropriately monotone.

*Proof.* We prove by induction on that implies . The cases where or is a limit ordinal are completely trivial, so consider the successor ordinal case. Suppose . Then there is some such that and . As and forks over , it certainly forks over . So implies , completing the inductive step. (I guess we only used induction in the limit ordinal case.) QED

Fact 2: If and , then . Non-forking extensions have the same rank.

*Proof.* In light of Fact 1, we only need to show that . We prove by induction on that implies . As before, the cases where is zero or a limit ordinal are completely trivial. Consider the successor ordinal case: . Then there exists such that and . We may move by an automorphism over so that . Since , it follows by left transitivity that , which in turn implies . Since , the inductive hypothesis implies . If , the fact that would imply by right-transitivity that , contradicting the choice of . So . Therefore implies , by definition of Lascar rank. This completes the inductive step, and the proof. QED

Fact 3: If are tuples, then .

*Proof.* We show by induction on that implies . As before, the only non-trivial case is the successor ordinal case. Suppose . Then there is with and . By induction, . By monotonicity of forking, , so implies , completing the inductive step. QED

For the left-hand side, we prove by induction on that if , then . For the base case , note that by Facts 1 and 3 above. For the successor case, suppose that . Then by definition of Lascar rank, there is some such that and . Move by an automorphism over so that . This implies that , by Fact 2.

By the inductive hypothesis, Because , monotonicity of implies that , that is, is a forking extension of . Consequently, completing the inductive step, in the case of successor ordinals. Finally in the limit ordinal case, suppose that is a limit ordinal, and . Then for every . By the inductive hypothesis, for every . Since ordinal addition is continuous in the right operand, holds, completing the inductive step in the limit ordinal case. We have shown that for every , Taking (or taking arbitrarily large when ), we conclude that the left-hand side of the Lascar inequalities holds.

For the right-hand side, we prove by induction on that The case is trivial, since Lascar rank is always at least 0. The limit ordinal case is also completely trivial. So consider the successor ordinal case: suppose that . Then by definition of Lascar rank, there exists such that , and . By the inductive hypothesis, If and , then by left-transitivity of forking independence, , a contradiction. So or . In the first case, Since by monotonicity of Lascar-rank (Fact 1), we see that completing the inductive step. In the second case, , so so we similarly have Either way, the inductive step holds. This completes the induction on . Now setting , we get the right-hand side of the Lascar inequalities.

Finally, we prove that if , then . We have already seen that so we only need to show that . We prove by joint induction on and that If or is 0, this follows from the fact that and . Otherwise, recall that the natural sum is the smallest ordinal greater than all ordinals of the form and for and . So, we need to show that if or , then or , respectively. We handle the first case; the second is completely symmetric. Since , it follows that , so there is some with and . Moving over , we may assume that . But then since , it follows by left-transitivity that , and in particular and . Since , we have (Fact 2). And since , , and , the inductive hypothesis ensures that . Now , since forks with , so This completes the proof. ::: ::: :::