Computability and Logic

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Computability and Logic
Slides by Prof.Yuxi Fu
BASICS, Department of Computer Science
Shanghai Jiao Tong University
April 13, 2015
Decidability, Undecidability, Partial Decidability
Decidability and Undecidability
A predicate M(x) is decidable if its characteristic function cM (x)
given by
1, if M(x) holds,
cM (x) =
0, if M(x) does not hold.
is computable.
The predicate M(x) is undecidable if it is not decidable.
Undecidability Result
Theorem. The problem ‘x ∈ Wx ’ is undecidable.
Proof. The characteristic function of this problem is given by
1, if x ∈ Wx ,
c(x) =
0, if x ∈
/ Wx .
Suppose c(x) was computable. Then the function g(x) defined
below would also be computable.
0,
if c(x) = 0,
g(x) =
undefined, if c(x) = 1.
Let m be an index for g, i.e. g = φm . Then
m ∈ Wm iff c(m) = 1 iff m ∈
/ Wm .
Corollary. There is a computable function h such that both
‘x ∈ Dom(h)’ and ‘x ∈ Ran(h)’ are undecidable.
Proof. Let
h(x) =
x,
if x ∈ Wx ,
undefined, if x ∈
/ Wx .
Clearly x ∈ Dom(h) iff x ∈ Wx iff x ∈ Ran(h).
Undecidability Result (Halting Problem)
Theorem. The problem ‘φx (y ) is defined’ is undecidable.
Proof. If y ∈ Wx were decidable then x ∈ Wx would be
decidable.
In this proof we have reduced the problem ‘x ∈ Wx ’ to the
problem ‘y ∈ Wx ’. The reduction shows that the latter is at least
as hard as the former.
Reduce
x ∈ Wx is reduced to the Halting problem.
Problem A is reduced to Problem B :
from the solution of Problem B we can effectively construct the
solution of Problem A.
Informally, Problem B is at least as hard as Problem A
Symbolically, A ≤ B.
Theorem. The problem ‘φx = 0’ is undecidable.
Proof. Consider the function f defined by
0,
if x ∈ Wx ,
f (x, y ) =
undefined, if x ∈
/ Wx .
By s-m-n theorem there is some total computable function k(x)
such that φk (x) (y) ' f (x, y).
It is clear that φk(x) = 0 iff x ∈ Wx .
Corollary. The problem ‘φx = φy ’ is undecidable.
Theorem. Let c be any number. The followings are
undecidable.
(a) Acceptance Problem: ‘c ∈ Wx ’,
(b) Printing Problem: ‘c ∈ Ex ’.
Proof. Consider the function f defined by
y,
if x ∈ Wx ,
f (x, y ) =
undefined, if x ∈
/ Wx .
It is clear that c ∈ Wk(x) iff x ∈ Wx iff c ∈ Ek(x) .
Rice’s Theorem
Theorem. (Rice)
Suppose ∅ ( B ( C1 . Then the problem ‘φx ∈ B’ is undecidable.
Proof. Suppose f∅ 6∈ B and g ∈ B. Let the function f be defined
by
g(y ),
if x ∈ Wx ,
f (x, y ) =
undefined, if x ∈
/ Wx .
It is clear that φk(x) ∈ B iff φk(x) = g iff x ∈ Wx .
(Un)decidable problems in other areas
1
The word problem for groups.
2
Diophantine equations: p(x1 , x2 , . . . , xn ) = 0 with integer
coefficients.
Hilbert’s tenth problem (1900): whether there is an
effective procedure that will determine whether any given
diophantine equation has a solution?
NO! (Matiyasevich 1970)
Mathematical logic
3
Validity is decidable in the propositional calculus.
Validity is undecidable in the first-order calculus.
Partially Decidable Predicates
A predicate M(x) of natural numbers is partially decidable if the
function given by
1,
if M(x) holds,
f (x) =
undefined, if M(x) does not hold,
is computable.
The function f (x) is the partial characteristic function.
Examples
1. The halting problem is partially decidable. Its partial
characteristic function is given by
1,
if Px (y) ↓,
f (x, y ) =
undefined, otherwise.
2. Any decidable predicate is partially decidable: make the
decision procedure to enter a loop whenever it outputs 0.
Examples
3. g(x) is a computable function. ‘x ∈ Dom(g)‘ is partially
decidable.
4. The problem ‘x ∈
/ Wx ’ is not partially decidable. The domain
of its partial characteristic function differs from the domain of
every computable function.
Theorem. A predicate M(x) is partially decidable iff there is a
computable function g(x) such that M(x) ⇔ x ∈ Dom(g).
Proof. g is essentially the partial characteristic function.
Theorem. A predicate M(x) is partially decidable iff there is a
decidable predicate R(x, y ) such that M(x) ⇔ ∃y.R(x, y).
Proof. If R(x, y) is decidable and M(x) ⇔ ∃y.R(x, y ), then
g(x) ' µyR(x, y) is computable. Clearly M(x) ⇔ x ∈ Dom(g).
Conversely suppose M(x) is partially decided by P. Let R(x, y )
be
R(x, y) ≡ P(x) ↓ in y steps.
Then R(x, y ) is decidable and M(x) ⇔ ∃yR(x, y).
Theorem. If M(x, y ) is partially decidable, so is ∃y.M(x, y).
Proof. Let R(x, y, z) be a decidable predicate such that
M(x, y) ⇔ ∃z.R(x, y, z). Then
∃y .M(x, y) ⇔ ∃u.R(x, (u)1 , (u)2 ).
where u = 2y 3z
Corollary. If M(x, y) is a partially decidable, where
y = (y1 , . . . , ym ), then so is ∃y1 . . . ∃ym M(x, y1 , . . . , ym ).
Examples
1. The following predicates are partially decidable
1
x ∈ Ey .
Proof. x ∈ Ey ⇔ ∃z∃t(Py (z) ↓ x in t steps)
2
Wx 6= ∅
Proof. Wx 6= ∅ ⇔ ∃y ∃t (Px (y) ↓ in t steps).
So Wx 6= ∅ is partially decidable.
Examples
2. Provability in predicate logic is partially decidable.
Theorem. A predicate M(x) is decidable iff both M(x) and
¬M(x) are partially decidable.
Proof.
Corollary. The problem ‘y ∈
/ Wx ’ is not partially decidable.
Theorem. Let f (x) be a partial function. Then f is computable
iff the predicate ‘f (x) ' y’ is partially decidable.
Proof. If f (x) is computable by P(x), then
f (x) ' y ⇔ ∃t.(P(x) ↓ y in t steps).
We are done by observing that ‘P(x) ↓ y in t steps’ is decidable.
Conversely let R(x, y , t) be such that
f (x) ' y ⇔ ∃t.R(x, y , t).
The equivalence gives rise to an algorithm.
Corollary
The predicate Px (y ) ↑ is not partially decidable.
Proof. O.W. The halting problem Px (y) ↓ would be decidable.

Computability and Logic

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