Is EQ CFG decidable?
Therefore EQCFG can not be decidable. Claim: EQCFG is co-Turing-recognizable. Proof: EQCFG is co-Turing-recognizable if EQCFG is Turing-recognizable. EQCFG = 1<G, H>|G is not a CFG, H is not a CFG, or both are CFGs and L(G) = L(H)l.
Is all CFG decidable?
ECFG = {〈G〉 | G is a CFG and L(G) = ∅} ECFG is a decidable language. Proof Sketch: Incrementally mark variables that can generate a sequence of terminals. Every CFG is decidable.
Why is a CFG decidable?
A language is decidable if there is a Turing Machine that halts and accepts strings that belong to the language, and halts and rejects strings that do not belong to the language.
Why Cannot we prove that Eqcfg is decidable?
No, EQCFG cannot be recognizable because we just argued that EQCFG is recognizable and hence if EQCFG was recognizable too, the language would be necessarily decidable (by Theorem 4.22 on page 183), but this is not the case. Consider the following instance P of Post correspondence problem.
Is first order logic decidable?
First-order logic is not decidable in general; in particular, the set of logical validities in any signature that includes equality and at least one other predicate with two or more arguments is not decidable. Logical systems extending first-order logic, such as second-order logic and type theory, are also undecidable.
Why is Allcfg undecidable?
Theorem: All CFG is undecidable. C1 is the starting configuration of M on w, Cℓ is an accepting configuration of M, Each Ci yields Ci+1 by transition function of M. A string is not an accepting computation history if it fails one or more of these conditions.
Is the empty set decidable?
The emptiness problem is undecidable for context-sensitive grammars, a fact that follows from the undecidability of the halting problem. It is, however, decidable for context-free grammars.
Is Empty CFG decidable?
What are decidable and undecidable problems?
A decision problem P is undecidable if the language L of all yes instances to P is not decidable. An undecidable language may be partially decidable but not decidable. Suppose, if a language is not even partially decidable, then there is no Turing machine that exists for the respective language.
How do you prove a language is decidable?
By definition, a language is decidable if there exists a Turing machine that accepts it, that is, halts on all inputs, and answers “Yes” on words in the language, “No” on words not in the language. Therefore one way of showing that a language is decidable is by describing a Turing machine that accepts it.
Why is predicate logic not decidable?
It’s hard to say what the “cause” is – mathematical phenomena have proofs, not causes. But the key reason for the undecidability is that predicate logic is too powerful; it’s powerful enough to describe the algorithm you might try to use, so it can circumvent it.
Why is fol not decidable?
Turing’s halting problem is used to show that there is no algorithm for determining whether B is logically entailed. First-order logic is complete because all entailed statements are provable, but is undecidable because there is no algorithm for deciding whether a given sentence is or is not logically entailed.
Why is ETM undecidable?
D rejects (D), but then H accepted (D,(D)) and hence D accepted (D), contradiction! So D cannot exist, so H cannot exist either (D was built from H). This means that ATM is undecidable. Theorem The language ATM is recognizable.
How do you prove Undecidability and reduction?
Lecture 40/65: Reducibility: A Technique for Proving Undecidability
What is decidable problem in TOC?
Decidable Problems
A problem is decidable if we can construct a Turing machine which will halt in finite amount of time for every input and give answer as ‘yes’ or ‘no’. A decidable problem has an algorithm to determine the answer for a given input.
How do you know if a language is decidable?
A language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w. Every decidable language is Turing-Acceptable. A decision problem P is decidable if the language L of all yes instances to P is decidable.
What is decidable problem example?
A problem is said to be Decidable if we can always construct a corresponding algorithm that can answer the problem correctly. We can intuitively understand Decidable problems by considering a simple example. Suppose we are asked to compute all the prime numbers in the range of 1000 to 2000.
Which is decidable problem?
Definition: A decision problem that can be solved by an algorithm that halts on all inputs in a finite number of steps. The associated language is called a decidable language. Also known as totally decidable problem, algorithmically solvable, recursively solvable.
How do you determine decidable?
How do you know if a problem is decidable?
A problem is said to be Decidable if we can always construct a corresponding algorithm that can answer the problem correctly.
Is modal logic decidable?
But unlike first-order logic, modal logic is decidable – showing fine-structure inside classical logic: with a delicate balance between expressive power and computational complexity.
Is first-order logic decidable?
Why ATM is not decidable?
D accepts <D> if H rejects <D,<D>> if D rejects <D> Either way: Contradiction! D cannot exist ⇒ H cannot exist Therefore, ATM is not a decidable language.
Is complement of ATM decidable?
Corollary 4.23: ATM is Turing-recognizable but not decidable, so its complement ATM is NOT Turing-recognizable.