Twierdzenbie Trachtenbrota
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Transcript Twierdzenbie Trachtenbrota
Trachtenbrotβs Theorem
The following problem is undecidable
Given: A sentence π of first order
logic
Question: Is π satisfiable in a finite
structure?
Theorem of Matyasevich
The following problem is undecidable
Given: A polynomial π π1 , π2 , β¦ , ππ of
multiple natural varibales and integer
coefficients
Ques: Does π have a zero, i.e., do there exist
natural numbers π1 , π2 , β¦ , ππ
such that π π1 , π2 , β¦ , ππ = 0?
Proof:
Reduction of Hilbertβs 10th problem to the
problem is a given first order sentence is
satisfiable in the finite.
The form of Hilbertβs problem:
Given: Two multivariate polynomials π1 , π2 of
degree π with all coefficients equal 1.
Question: Does the equation π1 = π2 have a
natural solution?
Reduction
β’ For each variable ππ in the equation we create
a unary relation symbol ππ
For each monomial
πΌπ = π1
π1
β
β¦ β
ππ
ππ
we use:
βA constant ππ
βAn π + 1-ary relation symbol π΄π
A sentence ππ corresponding to πΌπ
βπ₯0 π₯1 β¦ π₯π [π΄π (π₯0 , π₯1 , β¦ , π₯π ) β
(π1 π₯1 β§ β¦ β§ π1 π₯π1 β§
π2 π₯π1 +1 β§ β¦ β§ π2 π₯π1 +π2 β§
β¦
ππ π₯βππ βππ +1 β§ β¦ β§ ππ π₯βππ β§
π₯1+βππ = ππ β§
β¦
π₯π = ππ β§
π₯0 = ππ )]
That sentence enforces
π1
π΄π = |π1 |
ππ
β
β¦ β
|ππ |
And additionally the relation π΄π
contains only tuples beginning with ππ .
We add a 2π + 2-ary relation
symbol π΅ and a sentence Ξ²
ππ β ππ β§
πβ π
π1 β§ π2 β§ β― ππ β§
"π΅ is a 1-1 mapping
From the set of all tuples in relations π΄π related to π1
Onto the set of all tuples in relations π΄π related to π2 "