Proving properties of C(B) (proving properties of inductive
sets)
When we want to prove that every element of an inductive set has a certain property we
use a proof by induction, which is based on the Induction principle.
Induction Principle:
Let α = <U, f ini >i  I be a structure, and let B  U.
If S is a B-inductive subset of C(B) that is, B  S and is S closed in α, then S = C(B).
Proof of the induction principle:
To prove the induction principle, we need to show that S  C(B) and C(B)  S, therefore
(C(B) = S).
1) Show that S  C(B):
It is given in the induction principle that S is a B-inductive subset of C(B), therefore,
S  C(B).
2) Show that C(B)  S:
According to the definition of C(B), C(B) is the intersection of all S that are B-inductive
in α (C(B) =  {S: S is B-inductive in α}), therefore, C(B)  S, based on the definition of
intersection. In addition, we already proved that C(B) is the smallest B-inductive set in α,
therefore, it must be a subset of S.
Let P be a property.
To prove that P(x) holds for every element x in C(B) (or that every element x in C(B) has
property P), we use the Induction Principle and perform a proof by induction.
Theorem: For every x  C(B), P(x) holds.
This is the kind of statements we want to prove using induction
Let Sp = {x: x  C(B) and P(x) holds}.
Sp is the set consisting of all the elements x that are elements of C(B) and have a property
P.
To show that Sp = C(B), we need to show that Sp  C(B) and C(B)  Sp.
Sp consists of elements that are in C(B), therefore Sp is a subset of C(B) ( Sp  C(B)).
We already proved that C(B) is the smallest B-inductive set, therefore it must be a subset
of any other B-inductive set. If we can prove that Sp is B-inductive, then we also proved
that Sp is a subset of C(B) (C(B)  Sp)
So, following the Induction Principle, if we can prove that Sp is B-inductive and knowing
that Sp  C(B), then we can conclude that Sp and C(B) are the same set (Sp = C(B)).
Proof:
Let us prove that Sp is B-inductive. To do so, we perform the following steps, which we
call the schema of a proof by induction:
Schema of Proof by induction:
a) Base Case: Show that B is a subset of Sp (B  Sp), in other words, show that for every
element x in B, P(x) holds. Remember here, that the elements in B are defined by the base
rule of the inductive definition.
b) Inductive Step: Show that Sp is closed in α, i.e., for all i  I, if x1, …, xn i  Sp, then
f ini (x1, …, xn i )  Sp. In other words, show that if x1, …, xn i have property P, then f ini (x1,
…, xn i ) has property P as well.
To do so, we assume that x1, …, xn i  Sp (induction hypothesis), in other words that x1,
…, xn i have property P, and then we show that f ini (x1, …, xn i )  Sp, in other words that
f ini (x1, …, xn i ) has property P as well.
Let i  I be arbitrary and let x1, …, xn i  C(B).
Induction Hypothesis: x1, …, xn i  Sp (x1, …, xn i have property P).
Show that f ini (x1, …, xn i )  Sp (f ini (x1, …, xn i ) has property P).
Conclude that for all i  I, if x1, …, xn i have property P, then f ini (x1, …, xn i ) has
property P.
Example 1
Prove that every formula F has the following property P: F has an even number of
parentheses.
=> the proof is in your manuscript. Note that there is an error in that proof. In the proof in
your manuscript, on the Induction Step, you have:
The number of parentheses in F = the number of parentheses in G + the number of
parentheses in H.
=> The correct one is:
The number of parentheses in F = the number of parentheses in G + the number of
parentheses in H + 2.
Example 2
We define my-formulas as follows:
1) Base Clause: Every propositional sentence Ai (i=1,2,…) is a my-formula.
2) Inductive Clause:
(a) For every my-formula G, (  G} is also a my-formula.
(b) For every my-formulas G and H, {{G  H) is also a my-formula.
3) Final Rule: No expression in U* is a my-formula, unless it has to be one by 1., or
2. above.
Prove that for every my-formula F the following property P(F) holds: the number of “(”
in F plus the number of “{” in F, equals the number of “}” in F plus the number of “)” in
F plus the number of “  ” in F (i.e., n((F) + n{(F) = n)(F) + n}(F) + n  (F))
Proof by Induction:
Let F be an arbitrary my-formula.
(a) Base case:
Assume F is a propositional letter, i.e. F = Ai for some i{1,2,.....}. Then F has 0 (,
and 0 {, 0 }, 0 ), and 0  , so the number of ( plus the number if { is equal to the
number of ) plus the number of }, plus the number of  , i.e., n((F) + n{(F) = n)(F) +
n}(F) + n  (F).
Induction Step:
(b) Suppose F = (  G} and G is a my-formula.
Induction hypotheses: For my-formula G the property P(G) holds, i.e., the number of (
in G plus the number of { in G, is equal to the number of } in G plus the number of ) in
G plus the number of  in G, i.e., n((G) + n{(G) = n)(G) + n}(G) + n  (G).
n((F) = n((G) + 1
n{(F) = n{(G)
n)(F) = n)(G)
n}(F) = n}(G) + 1
n  (F) = n  (G).
Then, n((F) + n{(F) = n((G) + n{(G) + 1 =(by induction hypothesis) n)(G) + n}(G)
n  (G) + 1= n)(F) + n}(F) + n  (F)
Therefore P(F) holds (i.e., n((F) + n{(F) = n)(F) + n}(F) + n  (F) ).
(c) Suppose F = {{G  H), where G and H are my-formulas.
Induction hypotheses: For my-formula G the property P(G) holds, i.e., the number of (
in G plus the number of { in G, is equal to the number of } in G plus the number of ) in
G plus the number of  in G, i.e., n((G) + n{(G) = n)(G) + n}(G) + n  (G).
For my-formula H the property P(H) holds, i.e., the number of ( in H plus the number
of { in H, is equal to the number of } in H plus the number of ) in H plus the number of
 in H, i.e., n((H) + n{(H) = n)(H) + n}(H) + n  (H).
n((F) = n((G) + n((H)
n{(F) = n{(G) + n{(H) +2
n)(F) = n)(G) + n)(H) + 1
n}(F) = n}(G) + n}(H)
n  (F) = n  (G) + n  (H) + 1.
Then, n((F) + n{(F) = n((G) + n((H) + n{(G) + n{(G) + 2 =(by induction hypothesis)
n)(G) + n)(H) + n}(G) + n}(H) + n  (G) + n  (H) + 2= n)(F) + n}(F) + n  (F)
Therefore P(F) holds (i.e., n((F) + n{(F) = n)(F) + n}(F) + n  (F) ).
Since F was an arbitrary my-formula and we covered all the possibilities for the structure
of F, we proved the proposition.