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The question goes like this:

Ms. Li has just moved into a new home and plans to buy several potted plants at once. Her thoughts are as follows:

  1. Buy either Green Radish or Podocarpus.

  2. Buy at least one of Spider Plant, Green Radish, and Rubber Tree.

  3. Buy at least two of Podocarpus, Rubber Tree, and Sansevieria.

  4. If Podocarpus is bought, then Spider Plant is not bought.

If all the above conditions are met,

then Which of the following options A, B, C, D is necessarily correct?:

A. Ms. Li bought Sansevieria.

B. Ms. Li bought Rubber Tree.

C. Ms. Li bought either Podocarpus or Spider Plant.

D. Ms. Li bought at least three kinds of potted plants.


The reasoning goes like this:

The question asks which option is necessarily correct, and one approach is to start from a modus tollens reasoning.

The only if-then statement in the question is sentence 4: If Podocarpus is bought, then Spider Plant is not bought.

From sentence 1, we can derive that if Podocarpus is bought, then Green Radish is not bought either.

Combining these two conditions, we can conclude that if Podocarpus is bought, then both Spider Plant and Green Radish are not bought, which leads to the necessity of buying Rubber Tree.

From sentence 3, if Podocarpus is not bought, then Rubber Tree and Sansevieria must be bought.

Therefore, whether Podocarpus is bought or not, Rubber Tree must be bought. The answer is B.

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    $\begingroup$ Do you know the answer? Or how you would solve it by hand ? Or are you just looking for natural language driven AI solutions? $\endgroup$ Commented Feb 8, 2024 at 5:46
  • $\begingroup$ There is ambiguity in the constraint statements. Further, is it a statement that "all the following statements are true" or is meant to be a question of which is true (as allued to by @Syed)? Mathematica can deal with logical statements using Reduce and simplify using BooleanMinimize once statements have been coded. If it is a question about natural language processing: I defer to others. $\endgroup$ Commented Feb 8, 2024 at 6:53
  • $\begingroup$ BooleanMinimize[{G ⊻ P && S ∨ G ∨ R && (P ∧ R ) ∨ (P ∧ S) ∨ (R ∧ S) && P \[Implies] ¬ S}] gives {(! G && ! P) || ! S} $\endgroup$ Commented Feb 8, 2024 at 7:02
  • $\begingroup$ @Syed The question has been clarified. $\endgroup$ Commented Feb 8, 2024 at 7:27
  • $\begingroup$ @Syed I think your approach is what is desired. Your interpretation regards first statement as exclusive or (how do you know that), and you appear to use S for 2 different plants. $\endgroup$ Commented Feb 8, 2024 at 7:35

2 Answers 2

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Just a slight variation on the excellent answer given by Bill, which is too long for a comment.

Boolean Variables

The truth values of gr, sp, rt, s, pod indicate whether Green Radish, Spider Plant, Rubber Tree, Sansevieria, or Podcarpus are bought or not.

Entering Conditions

c1 = gr \[Xor] pod; c2 = sp \[Or] gr \[Or] rt; c3 = ( { pod, rt, s } // Query[ Total, Boole ] ) >= 2; c4 = pod \[Implies] \[Not] sp; conditions = And @@ { c1, c2, c3, c4 }; 

Note that you can enter the logical operators using Esc str Esc with str being xor, or, =>, or not.

Entering Options

options = Association[ "A" -> s, "B" -> rt, "C" -> pod \[Xor] sp, "D" -> ( { gr, rt, pod, sp, s } // Query[ Total, Boole ]) > 2 ]; 

Note that since we decided to understand "either ... or" as an exclusive or in encoding the conditions, we should be consistent in encoding option C.

Checking Necessity

A predicate P is necessary for the conditions C, if $C \implies P$ is $true$. Thus

options // Query[All /* Dataset, conditions \[Implies] #& /* FullSimplify /* SameAs[True] ] 

Solution

Remarks

  1. I find Query[Total, Boole] applied to a list of variables and then checking this sum against an integer easier than working with subsets and lots of alternatives.

  2. Unfortunately, BooleanMinimize seems to "choke" on sums of Booles (1.), but fortunately FullSimplify does not. Maybe someone can further comment on this observation?

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    $\begingroup$ +1. Thank you for the wonderful education on use of ‘Query’ for this. :) $\endgroup$ Commented Feb 8, 2024 at 14:23
  • $\begingroup$ @ubpdqn You are welcome. I must, in return, thank Seth Chandler for having instigated my more frequent use of it. $\endgroup$ Commented Feb 8, 2024 at 14:32
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This may be close.

cond1 = (gr && ! Pod) || (Pod && ! gr) cond2 = sp || gr || rt || (sp && gr) || (sp && rt) || (gr && rt) || (sp && gr && rt) cond3 = (Pod && rt) || (Pod && San) || (San && rt) || (Pod && rt && San) cond4 = Pod \[Implies] \[Not] sp cond = cond1 && cond2 && cond3 && cond4 BooleanMinimize[cond] (gr && ! Pod && rt && San) || (! gr && Pod && rt && ! sp) 

So she bought either

Green Radish and Rubber Tree and Sansevieria or

Podocarpus and Rubber Tree.

And the only necessary answer is B since the rubber tree is the only plant in both purchase options.

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    $\begingroup$ For the last step, you can use BooleanMinimize too e.g. Implies[(gr && ! Pod && rt && San) || (! gr && Pod && rt && ! sp), #] & /@ {San, rt, Pod || sp, Or @@ And @@@ Subsets[{gr, Pod, rt, San}, {3, 4}]} // BooleanMinimize $\endgroup$ Commented Feb 8, 2024 at 9:06
  • $\begingroup$ @xzczd Good to know. There are probably better ways to code some of the conditions than I have here also. I am a little rusty on my boolean logic. $\endgroup$ Commented Feb 9, 2024 at 0:48

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