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The instructor for my class assigned homework that is not covered in both the class and the textbook. As a result I found this problem and cannot find anyway to solve it. Can someone link some reading materials or explain the concept I'm missing please? I don't know what I'm looking at here.

Enter T or F depending on whether the formula is ambiguous (T) or not (F). (You must enter T or F -- True (true) and False (false) will not work.)

  1. p∨(q∨r)

  2. p∧q∧r

  3. p
  4. p∨q
  5. p∧q∨(q∧r)

This is homework, so I'm not asking anyone to do it for me per se, but an example would help me (and my equally confused classmates) out greatly!

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    It seems that you have to "flag" the formulae with T or F (the instruction between parentheses means that you are working in a computer environment and that ou **must** use T and F and not other strings... I think). Is it so ? – 2017-01-26
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    If so, you must flag the formulae that are "ambiguous". What is the def of "ambiguous" ? $p$ is not (on any imaginabel sense of "ambiguous"). $p∧q∧r$ ... maybe, meaning that we can read it either as $(p∧q)∧r$ or as $p∧(q∧r)$. – 2017-01-26
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    So ambiguity means it can be formed in more than one way? – 2017-01-26
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    "ambiguity" means : it can have different meanings. So, yes: in this case, **I think**, it means "can be formed in different ways". – 2017-01-26
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    Can't that be said of any equation? i.e. p = ~(~p) – 2017-01-26
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    NO; there is only one way to "read" it. – 2017-01-26

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I am not sure, but I imagine in that case "ambiguous" is understood as "can be interpreted as different nonequivalent formulas by adding parentheses". For instance, if I understand correctly, $p \wedge q \wedge r$ is not ambiguous, because the two ways to understand it, namely $p \wedge (q \wedge r)$ and $(p \wedge q) \wedge r$, are equivalent (they are either both true or both false, depending on whether $p$, $q$, and $r$ are).

The last one, however, would be ambiguous, because it can be understood either as $(p \wedge q) \vee (q \wedge r)$ or as $p \wedge (q \vee (q \wedge r))$, which are not equivalent. Indeed, if $p$ is false but $q$ and $r$ are true, the first one is true while the second one is false.

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    Unless the logical connectors have been assigned a priority. – 2017-01-26
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    Indeed. From the way the problem is phrased I assumed they have not. But I may be wrong. – 2017-01-26
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    This is just "instructor-dependent". :-) – 2017-01-26
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    The instructor finally responded to our pleas, you are correct. Invisible +1 – 2017-01-27
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    Only difference is it doesn't need to be non-equivalent. – 2017-01-27
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    Thank you for the feedback! – 2017-01-27