The definitions of \(f\) being dominated and bounded above by \(g\) only differ in the quantifier before \(k\) at the very start (universal for the hypothesis, existential for the goal), so it suffices to use any positive value for \(k\). We can use \(1\). The desired result then follows directly.

The proof is entirely analogous to the previous proof.

Both directions follow directly from the definition of asymptotic boundedness.

Suppose \(f\) is asymptotically bounded below by \(g\) and also asymptotically dominated by \(g\). We need to find a contradiction to prove the statement.

First, we claim that there exists \(x\_ 0\) such that for all \(x \ge x\_ 0\) we have \(f(x) \ge k \cdot g(x)\) for some \(k {\gt} 0\), \(g(x) \ge 0\) and \(f(x) \le k' \cdot g(x)\) for all \(k' {\gt} 0\). Each of the asymptotic assumption gives one such constant, so taking the maximum of all three gives the needed value.

For the contradiction, consider \(f(x) \le k' \cdot g(x)\) when \(k' = k / 2\). In this case, we have \(f(x) \le (k / 2) \cdot g(x)\), but we also have \(k \cdot g(x) \le f(x)\), leading by transitivity to \(k \cdot g(x) \le (k / 2) \cdot g(x)\), an obvious contradiction to the fact that \((k / 2) \cdot g(x) \le k \cdot g(x)\).

This is a direct application of Lemma 38.

This is a direct application of Lemma 38.

The proof is analogous to the proof of Lemma 38. This time, we set \(k' = k + 1\) and thus produce the false inequality \((k + 1) \cdot g(x) \le k \cdot g(x)\).

This is a direct application of Lemma 41.

This is a direct application of Lemma 41.

Suppose \(f\) both dominates \(g\) and is dominated by \(g\). Our goal is to find a contradiction. We have by definition the inequalities \(g(x) {\gt} 0\), \(f(x) \ge k\_ 1 \cdot g(x)\) and \(f(x) \le k\_ 2 \cdot g(x)\) for all \(k\_ 1 {\gt} 0\), \(k\_ 2 {\gt} 0\) and for all \(x \ge x\_ 0\) for some \(x\_ 0\). Note that we use the same constant \(x\_ 0\) for all inequalities with no loss of generality. In fact, we shall also use the same \(x\_ 0\) in the asymptotic positivity condition \(g(x) {\gt} 0\).

Fix \(k\_ 1 = 2\) and \(k\_ 2 = 1\) (generally, we only need \(k\_ 1 \ge k\_ 2\)), so we now have \(f(x) \ge 2 \cdot g(x)\) and \(f(x) \le g(x)\). From these inequalities, it immediately follows that \(2 \cdot g(x) \le g(x)\). However, since \(1 \le 2\) and \(g(x) {\gt} 0\), we have \(g(x) {\lt} 2 \cdot g(x)\). We thus have two contradicting inequalities, finishing the proof.

This is a direct application of Lemma 44.

This is a direct application of Lemma 44.

Proving asymptotic boundedness is equivalent to proving boundedness above and below. Both can be proved the same way - we choose \(1_K\) for \(K\) and \(1_R\) for \(N\), then the required asymptotic growth properties follow from definitions of identity elements for \(K\) and \(R\).

This follows directly from 47.

This follows directly from 47.

Let \(k\_ 1\) and \(k\_ 2\) be the constants such that \(f(x) \le k\_ 1 \cdot g(x)\) and \(g(x) \le k\_ 2 \cdot h(x)\) for sufficiently large \(x\). By multiplicativity and transitivity, we have \(f(x) \le k\_ 1 \cdot k\_ 2 \cdot h(x)\).

Let \(k\_ 1\) and \(k\_ 2\) be the constants such that \(f(x) \ge k\_ 1 \cdot g(x)\) and \(g(x) \ge k\_ 2 \cdot h(x)\) for sufficiently large \(x\). By multiplicativity and transitivity, we have \(f(x) \ge k\_ 1 \cdot k\_ 2 \cdot h(x)\).

A direct consequence of Lemma 50 and Lemma 51.

Let \(k {\gt} 0\). Then by first assumption, we have \(f(x) \le k \cdot g(x)\) for large \(x\). By the second assumption, we have \(g(x) \le h(x)\) for large \(x\). By multiplicativity and transitivity, we get \(f(x) \le k \cdot h(x)\).

Let \(k {\gt} 0\). Then by first assumption, we have \(f(x) \ge k \cdot g(x)\) for large \(x\). By the second assumption, we have \(g(x) \ge h(x)\) for large \(x\). By multiplicativity and transitivity, we get \(f(x) \ge k \cdot h(x)\).

Let \(k\) be the constant such that \(f\) is asymptotically less than \(k \cdot g\). By positive scalar multiplicativity of asymptotic inequality, \(c \cdot f\) is asymptotically less than \(c \cdot k \cdot g\). Since \(c \cdot k {\gt} 0\), this implies that \(c \cdot f\) is bounded above by \(g\).

Let \(k\) be the constant such that \(f\) is asymptotically greater than \(k \cdot g\). By positive scalar multiplicativity of asymptotic inequality, \(c \cdot f\) is asymptotically greater than \(c \cdot k \cdot g\). Since \(c \cdot k {\gt} 0\), this implies that \(c \cdot f\) is bounded below by \(g\).

Above boundedness is exactly Lemma 55 and below boundedness is exactly Lemma 56.

Let \(k\) be the constant such that \(f\) is asymptotically less than \(k \cdot g\). By positive scalar multiplicativity of asymptotic inequality, \(-c \cdot f\) is asymptotically less than \(-c \cdot k \cdot g\). Since \(-c \cdot k {\gt} 0\), this implies that \(c \cdot f\) is bounded above by \(-g\).

Let \(k\) be the constant such that \(f\) is asymptotically greater than \(k \cdot g\). By positive scalar multiplicativity of asymptotic inequality, \(-c \cdot f\) is asymptotically greater than \(-c \cdot k \cdot g\). Since \(-c \cdot k {\gt} 0\), this implies that \(c \cdot f\) is bounded below by \(-g\).

Above boundedness is exactly Lemma 58 and below boundedness is exactly Lemma 59.

Let \(k\_ 1\) and \(k\_ 2\) be the constants such that \(f\_ 1\) is asymptotically less than \(k\_ 1 \cdot g\) and \(f\_ 2\) is asymptotically less than \(k\_ 2 \cdot g\). By additivity of asymptotic inequality, \(f\_ 1 + f\_ 2\) is asymptotically less than \((k\_ 1 * k\_ 2) \cdot g\). It directly follows that \(f\_ 1 + f\_ 2\) is asymptotically bounded above by \(g\).

Let \(k\_ 1\) and \(k\_ 2\) be the constants such that \(f\_ 1\) is asymptotically greater than \(k\_ 1 \cdot g\) and \(f\_ 2\) is asymptotically greater than \(k\_ 2 \cdot g\). By additivity of asymptotic inequality, \(f\_ 1 + f\_ 2\) is asymptotically greater than \((k\_ 1 * k\_ 2) \cdot g\). It directly follows that \(f\_ 1 + f\_ 2\) is asymptotically bounded below by \(g\).

This is proved directly by Lemma 61 and Lemma 62.

This property immediately follows from Lemma 22.

This property immediately follows from Lemma 23.

This property immediately follows from Lemma 61 and Lemma 64.

This property immediately follows from Lemma 62 and Lemma 65.

We prove this with an application of Lemma 66 on Lemma 35.

We prove this with an application of Lemma 67 on Lemma 36.

Let \(k\_ 1\) and \(k\_ 2\) be the constants such that \(f\_ 1\) is asymptotically less than \(g\_ 1\) and \(f\_ 2\) is asymptotically less than \(g\_ 2\). By multiplicativity of asymptotic inequality, \(f\_ 1 * f\_ 2\) is asymptotically less than \(k\_ 1 \cdot g\_ 1 * k\_ 2 \cdot g\_ 2\), which is equivalent to asymptotic above boundedness of \(f\_ 1 * f\_ 2\) by \(g\_ 1 * g\_ 2\).

Let \(k\_ 1\) and \(k\_ 2\) be the constants such that \(f\_ 1\) is asymptotically greaterthan \(g\_ 1\) and \(f\_ 2\) is asymptotically greater than \(g\_ 2\). By multiplicativity of asymptotic inequality, \(f\_ 1 * f\_ 2\) is asymptotically less than \(k\_ 1 \cdot g\_ 1 * k\_ 2 \cdot g\_ 2\), which is equivalent to asymptotic below boundedness of \(f\_ 1 * f\_ 2\) by \(g\_ 1 * g\_ 2\).