MATE 3120 (LÓGICA 1)
PRACTICE PROBLEMS
JOHN GOODRICK
These problems are just for practice – they will not be graded. Solutions to
problems 1, 3, and 4 appear at the end.
1. In each of the formulas below, determine which variables occur as free
variables. Assume that the symbol set S contains a binary relation symbol R
and a ternary (“3-ary”) relation symbol Q.
ϕ = ∀v0 ∀v1 (Rv0 v1 → ¬Qv1 v2 v3 )
ψ = ∀v0 (v0 ≡ v1 → ∃v1 (Qv1 v2 v1 ∧ Rv0 v1 ))
2. (a) Suppose that S = {R, Q} with R a binary relation symbol and Q a
ternary relation symbol. Let A be the S-structure with universe N,
RA = (n, m) ∈ N2 : n < m ,
and
QA = (a, b, c) ∈ N3 : a + b = c .
Let β : {v0 , v1 , v2 , . . .} be any variable assignment such that β(v1 ) = 2 and
β(v2 ) = 0. Show that for the formula ψ in problem 1,
(A, β) |= ψ.
(b) Find an S-interpretation which satisfies ϕ in problem 1, and find another
which does not satisfy ϕ in problem 1.
3. For a term t in first order logic, let F (t) be the number of symbols
in t which are function symbols and let G(t) be the number of symbols in t
which are either variables or constant symbols. (Note that we count each time
a symbol occurs in t; so for instance, if f is a binary function symbol and
t = f f v0 v0 f v1 v0 , then F (t) = 3 since the symbol “f ” appears three times, and
G(t) = 4 since v0 is counted three times.)
Suppose that the symbol set S contains no unary function symbols (but
it might contain binary function symbols, constant symbols, etc.). Prove by
induction on terms that for any S-term t, G(t) > F (t).
4. For a first-order formula ϕ, let P (ϕ) be the number of symbols in ϕ which
are the left parenthesis “(”, and let C(ϕ) be the number of symbols in ϕ which
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JOHN GOODRICK
are one of the binary connectives ←, ∧, ∨, or ↔. (For example, for the ψ in
problem 1 above, P (ψ) = 2 and C(ψ) = 2.)
Prove by induction on formulas that for any first-order formula ϕ, P (ϕ) =
C(ϕ).
5. Suppose that S = {P0 , P1 , P2 , . . .} is a symbol set with one unary relation
symbol Pi for every natural number i. Given a formula ϕ in sentential logic
with sentential variable symbols belonging to the set {p0 , p1 , p2 , . . .}, we define
a first-order formula ϕ∗ from ϕ by recursion:
If ϕ = pi , then ϕ∗ = ∀v0 Pi v0 ;
If ϕ = nψ, then ϕ∗ = ¬ψ ∗ ;
If ϕ = cϕ1 ϕ2 (for some sentential formulas ϕ1 and ϕ2 ), then ϕ∗ = (ϕ∗1 → ϕ∗2 ).
Prove that for any sentential formula ϕ, |= ϕ if and only if |= ϕ∗ .
6. (a) Show that for any first-order formulas ϕ and ψ,
∀x (ϕ ∧ ψ)
is logically equivalent to
(∀xϕ ∧ ∀xψ)
and that
∃x (ϕ ∨ ψ)
is logically equivalent to
(∃xϕ ∨ ∃xψ) .
(b) Show that
∀x (ϕ ∨ ψ)
is not always equivalent to
(∀xϕ ∨ ∀xψ) ,
and that
∃x (ϕ ∧ ψ)
is not always equivalent to
(∃xϕ ∧ ∃xψ) .
(HINT: In both cases, it is sufficient to find one counterexample. To do
this, you will need to pick a symbol set S and pick specific formulas ϕ and ψ.
Also, in both cases, one of the formulas logically implies the other, but not vice
versa.)
MATE 3120 (LÓGICA 1)
PRACTICE PROBLEMS
3
SOLUTIONS
1. In ϕ, only the variables v2 and v3 occur free. (The variables v0 and v1 are
bound by the quantifiers ∀v0 and ∀v1 .)
In ψ, the variables v1 and v2 occur free and the variable v0 is bound. This
example is more confusing because the variable v1 occurs both freely and
bound (by the quantifier ∃v1 ), but we still say that v1 is a free variable of ψ.
3. The base cases are where t = x for a variable x or when t = c for a
constant symbol c. In either case t = x or t = c), we have that F (t) = 0 and
G(t) = 1, so it is true that G(t) > F (t).
For the induction step, suppose that t = f t0 t1 . . . tn−1 for some terms t0 , t1 , . . . , tn−1 ,
and assume that we already know that G(ti ) > F (ti ) for every i. Then
G(t) = G(f t0 . . . tn−1 ) = G(t0 ) + G(t1 ) + . . . + G(tn−1 )
and
F (t) = F (f t0 . . . tn−1 ) = 1 + F (t0 ) + . . . + F (tn−1 ).
By the induction hypothesis, G(ti ) ≥ F (ti ) + 1, and since S contains no unary
function symbols, n ≥ 2. Therefore
G(t0 ) + . . . + G(tn−1 ) ≥ F (t0 ) + . . . + F (tn−1 ) + n ≥ F (t0 ) + . . . + F (tn−1 ) + 2,
and so
G(t) = G(t0 ) + . . . + G(tn−1 ) > F (t0 ) + . . . + F (tn−1 ) + 1 = F (t).
This finishes the proof.
4. The base case is when ϕ is atomic. The first subcase is
ϕ = t0 ≡ t1
for some terms t0 and t1 . In this case, P (ϕ) = 0 and C(ϕ) = 0 (there are no
parentheses or connectives), so it is true that P (ϕ) = C(ϕ). Similarly, if
ϕ = Rt0 . . . tn−1
for some terms t0 , . . . , tn−1 and some n-ary relation symbol R, then we also
have that P (ϕ) = 0 = C(ϕ).
Now for the induction steps, we have to consider the cases:
(A) ϕ = ¬ψ,
(B) ϕ = (ϕ1 ? ϕ2 ) for some binary connective ? ∈ {→, ↔, ∧, ∨}, and
(C) ϕ = ∃xψ or ϕ = ∀xψ.
In the cases above, we are assuming that ψ, ϕ1 , and ϕ2 are formulas and that
we already know that P (ψ) = C(ψ), P (ϕ1 ) = C(ϕ1 ), and P (ϕ2 ) = C(ϕ2 ).
In case (A), P (ϕ) = P (¬ψ) = P (ψ) and C(ϕ) = C(¬ψ) = C(ψ). But by
induction hypothesis, P (ψ) = C(ψ), and so P (ϕ) = C(ϕ).
In case (B),
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P (ϕ) = P ((ϕ1 ? ϕ2 )) = 1 + P (ϕ1 ) + P (ϕ2 )
= 1 + C(ϕ1 ) + C(ϕ2 ), by induction hypothesis
= C((ϕ1 ? ϕ2 )) = C(ψ),
so again P (ϕ) = C(ϕ).
Case (C) has an identical proof as case (A).