6. Functions
Solutions
Exercises A.
1. Define a predicate is_function that takes as input a vector whose components are vectors of
dimension two. Then is_function has the following values:
is_function([]) = 1,
is_function([ [1, 2] ]) = 1,
is_function([ [1, 2], [3, 2] ]) = 1,
is_function([ [0, 5], [1, 4], [3, 0], [1, 4], [2, 5] ]) = 1, and
is_function([ [1, 2], [3, 2], [1, 4] ] = 0.
Test your function and write a description of the definition.
Hint: You may want to define an auxiliary function with input two vectors of dimension two.
What is important about such vectors when testing to see whether these vectors violate the
definition of function?
Solution
The DERIVE expressions below define the function is_function and provide tests for this
function and the auxiliary function test.
CaseMode := Sensitive
not(p) := 1 - p
test(v,w) := not(IF(v SUB 1 = w SUB 1 AND v SUB 2 /= w SUB 2))
test([1, 2], [1, 3])
test([1, 2], [3, 3])
is_function(v) := PRODUCT(PRODUCT(test(v SUB i, v SUB j), i, 1, j), j, 1,
DIMENSION(v))
[is_function([]), is_function([[1, 1]]), is_function([[1, 2], [1, 3]]),
is_function([[1, 2], [2, 1]])]
v := [[1, 6], [3, 5], [2, 6]]
w := [[1, 6], [3, 5], [1, 4], [2, 8]]
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[is_function(v), is_function(w)]
The function test determines whether or not two ordered pairs are acceptable in the definition of
a function as a set of ordered pairs. The function is_function tests every pair of ordered pairs in
its argument to determine if they violate the definition of function or not. The product of these
tests is 1 if and only if every pair is acceptable.
2. Define a function domain that computes the domain of a finite function. Some values of
domain are shown below:
domain([]) = [],
domain([ [5, 2] ]) = [5], and
domain([ [4, 2], [8, 3], [16, 4] ]) = [4, 8, 16].
Test your function and write a description of the definition.
Solution
The function domain defined below returns the first components of the vectors in a vector of
vectors.
domain(f) := VECTOR(f SUB i SUB 1, i, 1, DIMENSION(f))
[domain([]), domain([[1, 2]]), domain(v)]
3. Define a function range that computes the range of a finite function. Test your function and
write a description of the definition.
Solution
The function range returns the second components of the vectors in a vector of vectors.
range(f) := VECTOR(f SUB i SUB 2, i, 1, DIMENSION(f))
[range([]), range([[1, 2]]), range(v)]
4. Define a function restriction that computes the restriction of a function f to a subset of the
domain of f. Your function should return the values shown below:
f := [ [1, 1], [2, 4], [3, 9], [4, 16]],
restriction(f, []) = [],
restriction(f, [1, 2]) = [ [1, 1], [2, 4] ],
restriction(f, domain(f)) = f, and
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restriction(f, [0, 1]) = undefined.
Solution
Functions defined earlier can be used as auxiliary functions in solving this problem and later
problems. Open a file in which these function definitions have been saved.
FIRST(v) := v SUB 1
TAIL(v) := VECTOR(v SUB i, i, 2, DIMENSION(v))
CaseMode := Sensitive
in(a, v) := IF(DIMENSION(v) = 0, 0, IF(a = FIRST(v), 1, in(a, TAIL(v))))
set(v) := IF(DIMENSION(v) = 0, v, IF(in(FIRST(v), TAIL(v)) = 1, set(TAIL(v)),
APPEND([FIRST(v)], set(TAIL(v)))))
union(s, t) := set(APPEND(s, t))
delete(a, s) := IF(DIMENSION(s) = 0, s, IF(a = FIRST(s), delete(a, TAIL(s)),
APPEND([FIRST(s)], delete(a, TAIL(s)))))
inter(s, t) := set(IF(DIMENSION(s) = 0, s, IF(in(FIRST(s), t) = 1, APPEND([FIRST(s)],
inter(TAIL(s), t)), inter(TAIL(s), t))))
diff(s, t) := set(IF(DIMENSION(t) = 0, s, IF(in(FIRST(t), s) = 1,
diff(delete(FIRST(t), s), TAIL(t)), diff(s, TAIL(t)))))
is_subset(s, t) := PRODUCT(in(s SUB i, t), i, 1, DIMENSION(s))
and(p, q) := p*q
eq(s, t) := and(is_subset(s, t), is_subset(t, s))
disjoint(s, t) := eq(inter(s, t), [])
restrict(f, s) := IF(DIMENSION(f) = 0, [], IF(in((FIRST(f)) SUB 1, s) = 1,
union([FIRST(f)], restrict(TAIL(f), s)), restrict(TAIL(f),s )))
[restrict([], []), restrict(v, []), restrict(v, [1, 3]), restrict(v, domain(v))]
restriction(f, s) := IF(is_subset(s, domain(f))=1, restrict(f, s ), "The set s is not a subset of the
domain of f.")
[restriction(v, [2, 1]), restriction(v, [1, 4])]
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The recursive function restrict adds the first element of f to the restricted function’s set only
when the element’s first component is in the set s. The function restrict returns the restriction of
the function f to the set s when s is a subset of the domain of f. The function restriction checks
that s is a subset of the domain of f.
5. Determine which of the statements below are true and which are false. Prove any true
statement, and give a counterexample (using DERIVE) for any false statement. Let f: X  Y be
a function.
a. For all subsets A and B of X, if A  B, then f(A)  f(B).
Solution
The statement is true.
Proposition: For all sets X and Y, subsets A and B of X, and functions f: X  Y, if A  B, then
f(A)  f(B).
Proof: Suppose that X and Y are sets, A and B are subsets of X, f: X  Y is a function, and
A  B. Further assume that y  f(A). Then there exists a  A such that f(a) = y. Since A is a
subset of B, a  B. Thus f(x)  f(B). Therefore, f(A)  f(B). 
b. For all subsets A and B of X, f(A  B) = f(A)  f(B).
Solution
The counterexample below shows that the statement is false. Define f, A, and B, and then
simplify the next two expressions.
[f := [[1, 3], [2, 3]], A := [1, 2], B := [1]]
range(restriction(f, diff(A, B)))
diff(range(restriction(f, A)), range(restriction(f, B)))
c. For all subsets A and B of X, f(A  B) = f(A)  f(B).
Solution
The statement is false by the counterexample below.
[f := [[1, 5], [2, 6], [3, 5]], A := [1, 2], B := [2, 3]]
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range(restriction(f, inter(A, B)))
inter(range(restriction(f, A)), range(restriction(f, B)))
d. For all subsets A and B of X, f(A  B) = f(A)  f(B).
Solution
The statement is true.
Proposition. For all sets X and Y, subsets A and B of X and functions f: X  Y, f(A  B) =
f(A)  f(B).
Proof: Suppose that X and Y are sets, A and B are subsets of X, and f: X  Y is a function.
Then an element y  f(A  B) iff there exists x  A  B such that f(x) = y
iff f(x) = y for some x  A or f(x) = y for some x  B
iff y  f(A)  f(B).
Therefore, f(A  B) = f(A)  f(B). 
Exercises B.
1. Define predicates is_oneone, is_onto, and is_bijective that perform the indicated tests on
finite functions. Consider the functions:
f := [ [1, 6], [3, 5], [2, 6] ] and
g := [ [1, 5], [2, 6], [3, 12] ]
The predicates should yield the following values:
is_oneone( [] ) = 1,
is_oneone( f ) = 0,
is_oneone( g ) = 1,
is_onto( [], [] ) = 1
is_onto( [], [1, 2]) = 0
is_onto( f, [5, 6] ) = 1,
is_onto( f, [4, 5, 6] ) = 0
is_onto( g, [12, 5, 6] ) = 1
is_bijective( [], [] ) = 1,
is_bijective( f, [5, 6] ) = 0, and
is_bijective( g, [5, 6, 12] ) = 1.
Test your predicates and write descriptions of their definitions.
Solution
The first four DERIVE expressions below define the functions. The expressions that follow can
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be entered to define examples and to test these functions.
The function test2 is an auxiliary function that determines whether or not a pair of twodimensional vectors violates the definition of injection. The product of the results for all ordered
pairs in the definition of a function f is 1 if and only if there are no violations, and this product
provides the definition of the function is_oneone.
The functions is_onto and is_bijective directly implement the definitions of these properties.
test2(v, w) := not(IF(v SUB 2 = w SUB 2 AND v SUB 1 /= w SUB 1))
is_oneone(f) := PRODUCT(PRODUCT(test2(f SUB i, f SUB j), i, 1, j), j, 1, DIMENSION(f))
is_onto(f, s) := eq(range(f), s)
is_bijective(f, s) := and(is_onto(f, s), is_oneone(f))
g := [[1, 4], [2, 5], [3, 4]]
[domain(g), range(g), is_oneone(g), is_onto(g, [4, 5]), is_bijective(g, [4, 5])
[is_function(g), domain(g), range(g), is_oneone(g), is_onto(g,[4,5]), is_bijective(g, [4, 5])]
The function g with codomain [4,5] is onto but not one-to-one.
[domain(g), range(g), is_oneone(g), is_onto(g, [4, 5, 6]), is_bijective(g, [4, 5, 6])]
The function g with codomain [4,5,6] is not onto and not one-to-one.
h := [[1, 4], [2, 5], [3, 6]]
[domain(h), range(h), is_oneone(h), is_onto(h, [4, 5, 6]), is_bijective(h,[4, 5, 6])]
The function h with codomain [4,5,6] is onto and one-to-one.
2. Define a DERIVE function val that evaluates a finite function. For the finite functions f and
g of question #1, we want the following results:
val(f, 1) = 5,
val(f, 2) = 6,
val(f, 4) = undefined, and
val(g, 3) = 12.
Test your function val and write a description of the definition.
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Solution
The function val1 is an auxiliary function that computes the sum of those second components
where the first component equals n. The function val returns “undefined” if n is not in the
domain of the function f, and it removes multiple occurrences of vectors from the definition of f
by applying the function set to f. Except for the three function definitions, the other DERIVE
expressions below can be used to test the functions.
val1(f, n) := SUM(IF(f SUB i SUB 1 = n, f SUB i SUB 2, 0), i, 1, DIMENSION(f))
[val1(f, 3), val1(h, 3)]
val(f, n) := IF(in(n, domain(f)) = 1, val1(set(f), n), "undefined")
val(f, 3)
[val(f, 4), val(h, 4)]
Exercises C.
1. Define a functional inverse that returns the inverse function of a one-to-one finite function.
For example, the following values should be returned:
inverse([ [1, 5], [2, 6] ] ) = [ [5, 1], [6, 2] ],
inverse( [] ) = []
inverse( [[1, 5], [2, 5] ] ) = undefined.
Test your functional inverse and write a description of its definition.
Solution
The function inverse1 defined below is an auxiliary function that interchanges the two
components of every ordered pair in the definition of f. The function inverse checks that the
result is a function by checking that f is injective.
inverse1(f) := VECTOR([f SUB i SUB 2, f SUB i SUB 1], i, 1, DIMENSION(f))
inverse1(h)
inverse(f) := IF(is_oneone(f) = 1, inverse1(f), "undefined")
[inverse(g), inverse(h)]
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2. Define DERIVE functionals whose two arguments are finite functions (expressed as vectors
with vector components of two dimensions) with real values. The values of the functionals are
finite functions. The new DERIVE functionals to be defined are:
a. sum
b. product
c. composition
Here are some examples of finite functions and the values of the functionals evaluated on these
functions.
f := [ [1, 6], [2, 8], [3, 10] ],
g := [ [1, -2], [3, 0], [2, -3] ],
h := [ [6, 5], [10, 11], [4, 3], [8, 4] ],
sum(f, g) = [ [1, 4], [2, 5], [3, 10] ],
sum(f, h) = undefined,
product(f, g) = [ [1, -12], [2, -24], [3, 0] ]
product(g, h) = undefined,
composition(h, f) = [ [1, 5], [2, 4], [3, 11] ], and
composition(h, g) = undefined.
These functionals should test that the domains of the input functions are suitable. Test your
functionals and write a description of the definitions.
Solution
The functionals sum, product, and composition are direct translations of the mathematical
concepts to DERIVE expressions. In each case auxiliary functions are first defined to
implement the operation, and then checks for validity of arguments are included in the required
functions. The expressions that are not definitions can be simplified to test the functionals.
sum1(f, g) := VECTOR([f SUB i SUB 1, f SUB i SUB 2 + val(g, f SUB i SUB 1)], i, 1,
DIMENSION(f))
sum1(g, h)
sum(f, g) := IF(eq(domain(f), domain(g)) = 1, sum1(f, g),"undefined")
k := [[0, 4], [2, 10]]
[sum(h, k), sum(g, h), sum(h, g)]
product1(f, g) := VECTOR([f SUB i SUB 1, f SUB i SUB 2*val(g, f SUB i SUB 1)], i, 1,
DIMENSION(f))
product(f, g) := IF(eq(domain(f), domain(g)) = 1, product1(f, g), "undefined")
[product(h, k), product(g, h), product(h, g)]
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composition1(f, g) := VECTOR([g SUB i SUB 1, val(f, g SUB i SUB 2)] ,i, 1,
DIMENSION(g))
m := [[1, 2], [2, 3], [3, 3], [0, 1], [4, 2]]
[composition1(g, m), composition1(h, m)]
composition(f, g) := IF(is_subset(range(g), domain(f)) = 1, composition1(f,g ),
" undefined")
[composition(g, m), composition(h, m), composition(h, g)]
3. Determine which statements below are true and which are false. Prove any true statement,
and give a counterexample (using DERIVE) for any false statement. Let X and Y be sets.
a. If f: X  Y and g: X  Y are one-to-one functions, then so is their sum.
b. If f: X  Y and g: X  Y are one-to-one functions, then so is their product.
Solution
Statements a and b are shown to be false by the following counterexamples.
f := [[1, 0], [2, 5]]
g := [[1, 5], [2, 0]]
[and(is_oneone(f), is_oneone(g)), is_oneone(sum(f, g)), is_oneone(product(f, g))]
c. If the functions f: X  Y and g: X  Y are onto a set S, then so is their sum.
d. If the functions f: X  Y and g: X  Y are onto a set S, then so is their product.
Solution
Statements c and d are shown to be false by simplifying the DERIVE expression below.
[and(is_onto(f, [0, 5]), is_onto(g, [0, 5])), is_onto(sum(f, g), [0,5]),
is_onto(product(f, g), [0, 5])]
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e. If the composition fg is one-to-one, so is the function f.
Solution
The counterexample below shows that statement e is false.
f := [[2, 5], [3, 5]]
g := [[1, 2]]
[is_oneone(composition(f, g)), is_oneone(f)]
f. If the composition fg is one-to-one, then so is the function g.
Solution
Statement f is true.
Proposition: For all functions g: X  Y and f: Y  Z, if the composition fg is one-to-one, then
so is the function g.
Proof: Suppose that g:X  Y and f: Y  Z are functions and the composition fg is one-to-one.
Assume that g(a) = g(b) for some elements a and b in X. Then f(g(a)) = f(g(b)). Thus, since the
composition fg is one-to-one, a = b. Therefore, g is one-to-one. 
g. If the composition fg is onto a set S, then so is f.
Solution
Statement g is true.
Proposition: For all functions g: X  Y and f: Y  Z, if the composition fg is onto Z, then so
is the function f.
Proof: Suppose that g: X  Y and f: Y  Z are functions and the composition fg is onto Z.
Let z be an element of Z. Then there exists x in X such that f(g(x)) = z since fg is onto. Let y =
g(x). Then f(y) = f(g(x)) = z. Therefore f is onto Z. 
h. If the composition fg is onto range(fg), then g is onto domain(f).
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Solution
Since every function is onto its range, the hypothesis is true in every case. Simplify the DERIVE
expression to show that the functions f and g provide a counterexample.
[is_onto(composition(f, g), range(composition(f, g))), is_onto(g,domain(f))]
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