It's quite a short question: Refering to this Wikipedia article p U q means, that p has to hold at least until at some point q holds.

So, does it mean:

  1. On each path starting from the current state p has to hold in each state until a state is reached in which q holds.
  2. On each path starting from the current state p has to hold in any state before a state is reached in which q holds.

Example: Suppose p holds in s3 and q holds in s5. Let s0 be the starting state and the set of edges given by { (s0,s1),(s1,s2),(s2,s3),(s3,s4),(s4,s5) }. Let's assume p and q don't hold in all the other states. Is p U q satisfied, starting from s0?


1 Answer 1


$pUq$, as is, is not a well-formulated CTL (or more generally - CTL*) formula. A CTL* formula is a state formula, which starts with has a path quantifier ($A$ or $E$).

$pUq$ is a path formula, which means it refers to a single (linear) computation. A computation satisfies $pUq$ if $p$ holds in every state until $q$ holds (and $q$ must hold at some state).

Translating to CTL, you can talk about two different formulas: $ApUq$ and $EpUq$. The former is true in a state if every path starting from this state satisfies $pUq$. The latter is true in a state if there exists such a path starting from that state.

Does that clear things up?

  • $\begingroup$ Yes, it does. Thanks a lot. One further question: If I get the things write, if a state holds p U q then, p must hold in that state. Is that true? $\endgroup$
    – 0xbadf00d
    Feb 1, 2014 at 21:40
  • 1
    $\begingroup$ $pUq$ is a path formula, so I guess you mean something like $ApUq$ or $EpUq$. In that case - either $p$ or $q$ must hold in that state. If $q$ holds, then the eventuality is immediately fulfilled. $\endgroup$
    – Shaull
    Feb 2, 2014 at 6:43
  • $\begingroup$ So, A[p U q] and E[p U q] are fulfilled when q holds in the current state. Even if p doesn't hold in that state? Seems to make sense. Thanks again. $\endgroup$
    – 0xbadf00d
    Feb 2, 2014 at 9:44
  • $\begingroup$ That is correct. $\endgroup$
    – Shaull
    Feb 2, 2014 at 9:48

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