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1answer
44 views

Using diagonal argument to prove that $H(x) = \mu y T(x,x,y)$ has no total computable extension

Hello everyone just like the title says I want to prove that $H(x) = \mu y T(x,x,y)$ has no total computable extension such that if we had a function $BIG(x)$ that is both total and agrees with $H(x)$ ...
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2answers
113 views

The minimization operator is an effective operator

Assume $\{f_i^{(n)}\}_{i=0}^\infty$ is a Gödel enumeration of the $\mu$-recursive functions of $n$ arguments, such that the $S^m_n$ theorem and the universal function theorem hold. Denote the set of ...
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1answer
55 views

Techniques (tools) to convert temporal logic (CTL,CTL* or LTL) to μ-calculus formulae

Suppose one wants to use a μ-calculus model checker, but specify things in temporal logics, which is easier (more intuitive). Is there a technique (even better, a tool) that automatically translates ...
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3answers
270 views

Why use $\mu$-calculus and not LTL,CTL,CTL*?

It is known that the temporal logics LTL,CTL,CTL* can be translated/embedded into the $\mu$-calculus. In other words, the (modal) $\mu$-calculus subsumes these logics, (i.e. it is more expressive.) ...
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1answer
60 views

Have I got the right understanding of the mu operator?

I have a homework problem that says: For $g(x,y)=xy-5$ compute $h(x) = \mu y(g(x,y))$ and determine its domain. I was under the impression that this means the least y such that $g(x,y)=0$, so then ...
1
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1answer
154 views

Intuitive meaning of modal $\mu$-calculus formula

I am solving one of the past exams and I am not certain with my solution to one of the exercises. The exercise is asking to give intuitive meaning to modal $\mu$-calculus formula: $$ \phi = \mu Z. ...