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the statement is: if p, then q, unless r

I'm to convert it into a compound proposition.

I've reasoned that:

when p is true AND r is true, q is false
when p is true AND r is false, q is true
when p is false, we can't know if q is true

But, how to convert this into compound propositions?

$ (p \land \lnot r) \to q $
$ (p \land r) \to \lnot q $

How can we express both of these at the same time? or does each of them already contain the other?

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  • $\begingroup$ Are we to parse the statement as "if p then (q unless r)" or "(if p then q) unless r"? $\endgroup$ – user856 Feb 13 '14 at 22:56
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    $\begingroup$ @Rahul: Is a there a difference? I would have thought the only counterexample to each would be $p \land \lnot q \land \lnot r$ $\endgroup$ – Henry Feb 13 '14 at 23:02
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    $\begingroup$ Not sure what your book means by "unless." Nor whether it is "(if $p$ then $q$) unless $r$" or it is "if $p$ then ($q$ unless $r$)." My intuitions is that your first statements is wrong - you can't conclude either way about $q$ if $p$ and $r$ are true. But it really depends on what is meant by "unless." $\endgroup$ – Thomas Andrews Feb 13 '14 at 23:23
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    $\begingroup$ Elaborating on Thomas's point. You need to clear up the order of operations, i.e., what is the left operand of "unless." AND you need to be clear what "unless" means, because "unless" is not consistently defined in mathematics. "A unless B" is sometimes taken to mean "A or B", and it's sometimes taken to mean "A xor B." Compare "I won't go to the concert unless it's free." (A or B, since I still might not go, even if it's free.) and "He's at the airport unless someone picked him up." (A xor B, since if someone picked him up, he's not at the airport.) $\endgroup$ – Steve Kass Feb 14 '14 at 1:39
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    $\begingroup$ We have the suggestion of S.C.Kleene, Mathematical Logic (1967), [page 64] : " $A \lor B$ is $A$ unless $B$ [usually]". So, I propose to read "if p, then q, unless r" as $(p \rightarrow q) \lor r$. See also the following post. $\endgroup$ – Mauro ALLEGRANZA Feb 14 '14 at 18:32
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Start with a truth table

 p | q | r || (p & ~r) -> q |(p & r) -> ~q  ||  p, then q, unless r
---|---|---||----------m----|--------m------||-----------------------
 T | T | T ||    F F   T    |   T    F  F   ||           ?
 T | T | F ||    T T   F    |   F    T  F   ||           ?
 T | F | T ||    F F   T    |   T    T  T   ||           ?
 T | F | F ||    T T   F    |   F    T  T   ||           ?
 F | T | T ||    F F   T    |   F    T  F   ||           ?
 F | T | F ||    F T   T    |   F    T  F   ||           ?
 F | F | T ||    F F   T    |   F    T  T   ||           ?
 F | F | F ||    F T   T    |   F    T  T   ||           ?

First conclusion $ (p \land \lnot r) \to q $ and $ (p \land r) \to \lnot q $ are not identical.

Then what is the truth table of "p, then q, unless r" ?

That question you have to do yourself, and you will need to argue for each line without reference to the formula's

GOOD LUCK

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It's $\lnot r \rightarrow (p \rightarrow q) $ I think

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  • $\begingroup$ thanks, but what does the question mark mean? $\endgroup$ – compguy24 Feb 13 '14 at 23:02
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    $\begingroup$ Are you sure? Why not $(p \rightarrow q) \lor r$? $\endgroup$ – Henry Feb 13 '14 at 23:08
  • $\begingroup$ @Henry my bad.. Thanks for correcting me. $\endgroup$ – user76568 Feb 13 '14 at 23:14

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