Examples on Proofs by Induction
Example1:
Let U be the set of all strings over the alphabet {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q,
r, s, #, $}, i.e., U = {a, …, s, #, $}*.
We define my-words as follows:
1) Base Clause: Any single character of the set {a, …, s} is a my-word.
2) Inductive Clause:
(a) If F is a my-word, then #F# is also a my-word.
(b) If F and G are my-words, then G$H is also a my-word.
3) Final Clause: No element of U is a my-word unless it has to be one by 1) or
2) above.
Prove that for every my-word F the following property P(F) holds: the length of F is odd,
i.e., the number of characters in F is odd, where a character is any member of the set {a,
b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, #, $}.
Proof by Induction:
Let F be an arbitrary my-word.
(a) Base case:
Assume F is an element of the set {a, …, s}. Then, F has length one, which is odd.
Therefore, P(F) holds.
Induction Step:
(b) Suppose F = #G#, where G is a my-word.
Induction hypotheses: For my-word G the property P(G) holds, i.e., G has odd length.
The length of F equals to the length of G + 2 (we add two more elements to G, which
are the elements # #).
By the induction hypothesis, the length of G is odd. So, we add 2 to an odd number,
which results to another odd number.
Therefore, the length of F is odd, i.e., P(F) holds.
(c) Suppose F = G$H, where G and H are a my-words.
Induction hypotheses: For my-words G and H, P(G) and P(H) hold, i.e., G and H have
odd length.
The length of F equals to the length of G + length of H + 1 (we add one more element
to the length of G plus the length of H, which is the element $).
By the induction hypothesis, the length of G and the length of H are odd. So, we add
two odd numbers together, which results in an even number, and then, we add 1 to that
even number, which results in an odd number.
Therefore, the length of F is odd, i.e., P(F) holds.
Since F was an arbitrary my-word and we covered all the possibilities for the structure of
F, we proved the proposition.
Example2:
Let U be the set of all strings over the alphabet {  , {, },  , (, ), Ai: i  IN+}, i.e., U =
{  , {, },  , (, ), Ai: i  IN+}*.
We define my-formulas as follows:
1. Base Clause: Every propositional sentence Ai (i  IN+) is a my-formula.
2. Inductive Clause:
a. If G is a my-formula, then (  G} is also a my-formula.
b. If G and H are my-formulas, then {{G  H) is also my-formula.
3. Final Clause: No element of 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.
Example3:
Let U be the set of all strings over the alphabet {♥, ♦, ●, ■, ♣, Ai: i  IN+}, i.e., U = {♥,
♦, ●, ■, ♣, Ai: i  IN+}*.
We define draw-lines as follows:
1. Every Ai, where i  IN+ is a draw-line.
2. If G is a draw-line, then ♥G♦ is also a draw-line.
3. If G and H are draw-lines, then ●●G■H♣ is also a draw-line.
Prove that for every draw-line 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 draw-line.
(a) Base case:
Assume F = Ai for some i{1,2,.....}. Then F has 0 ♥, and 0 ●, 0 ♦, 0 ♣, and 0■, so
the number of ♥ plus the number of ● 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♦, where G is a draw-line.
Induction hypotheses: For draw-line 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) + 1
n♣(F) = n♣(G)
n■(F) = n■(G).
Then, n♥(F) + n●(F) = n♥(G) + 1 + n●(G) (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 draw-lines.
Induction hypotheses: For draw-line 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 draw-line 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)
n♣(F) = n♣(G) + n♣(H) + 1
n■(F) = n■(G) + n■(H) + 1.
Then, n♥(F) + n●(F) = n♥(G) + n♥(H) + n●(G) + n●(H) + 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 draw-line and we covered all the possibilities for the structure of
F, we proved the proposition.
Example4:
Let U be the set of all strings over the alphabet {♥, ♦, ☺, ●, ■, ♣}, i.e., U = {♥, ♦, ☺, ●,
■, ♣}*.
We define drawing-figures as follows:
1. Base Clause: Every member of the set {♥, ☺, ■} is a drawing-figure.
2. Inductive Clause:
a. If G is a drawing-figure, then ♥☺G■ is also a drawing-figure.
b. If G and H are drawing-figures, then ●G♦H♣♣ is also a drawing-figure.
3. Final Clause: No element of U is a drawing figure unless it has to be one by 1. or
2. above.
Prove that for every drawing-figure 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 drawing-figure.
(a) Base case: Assume F is a member of the set {♥, ☺, ■}.
Then F has 0 ♥, 0 ♣, 0 ●, 0 ♦,and 0☺, so the number of ♥ in F plus the number of ♣
in F is equal to 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).
Induction Step:
(b) Suppose F = ♥☺G■, where G is a drawing-figure.
Induction hypotheses: For drawing-figure 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)
n☺(F) = n☺(G) + 1.
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 drawing-figures.
Induction hypotheses: For drawing-figure 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 drawing-figure 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) + 1
n☺(F) = n☺(G) + n☺(H).
Then, n♥(F) + n♣(F) = n♥(G) + n♣(G) + n♥(H) + n♣(H) + 2 =(by induction hypothesis)
n●(G) + n♦(G) + n☺(G) + n●(H) + n♦(H) + 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 drawing-figure and we covered all the possibilities for the
structure of F, we proved the proposition.