Theorem: There exist expressions with three constants.
Reasonable proof: Consider (1+2)+3 and the definition of expressions.
More pedenatic proof:
Consider (1+2)+3 and the definition of E. Since E_3 is subset of E,
it suffices to show (1+2)+3 is in E_3. By definition of E_3,
it suffices to show that (1+2) is in E_2 and 3 is in E_2.
To show (1+2) is in E_2, it suffices to show 1 is in E_1 and
2 is in E_1.
1 is in E_1 by definition of E_1 (includes all constants).
2 is in E_1 by definition of E_1 (includes all constants).
3 is in E_2 (includes all constants).
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Theorem: All expressions contain a constant or variable.
Proof #1 (induction on height):
By induction on the height of an arbitrary expression e.
Base case: Height h = 0
If an expression has height 0, it must be a variable or a constant.
Therefore it contains one.
Inductive case: Height h > 0 If an expression e has height h > 0,
then it is either e1+e2 or e1*e2 for some e1 and e2. In either
case, e1 has height at most h-1. So by induction e1 contains a
constant or variable. Therefore, e also contains a constant or
variable.
Proof #2 (structural induction):
By induction on the structure of an arbitrary expression e.
There are four cases:
If e is a variable, then it contains a variable or constant.
If e is a constant, then it contains a variable or constant.
If e is e1+e2 for some e1 and e2, then by induction e1 contains a variable
or constant. Therefore, e also contains a constant or variable.
If e is e1*e2 for some e1 and e2, then by induction e1 contains a variable
or constant. Therefore, e also contains a constant or variable.