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Notty Logic

Age 16 to 18
Challenge Level Yellow star
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Patrick from Woodbridge school sent us his thoughts, which were artfully clear, in which he reduced each sentence to a clear logical statement which could then be NOTted:

Using de Morgan's law that NOT(A and B) = NOT(A) OR NOT(B) and assuming that Bad means NOT(GOOD):

1. A good pet is friendly and furry is equivalent to

GOOD = Friendly AND Furry

so

NOT GOOD = NOT(Friendly AND Furry)

Using de Morgan's law

NOT GOOD = NOT(friendly) OR NOT(furry)

so d) A bad pet (NOT good) is unfriendly (NOT friendly) OR unfurry (NOT furry).

2. That man is lying or I'll eat my hat

Let X = Lying OR hat

NOT(X)

= NOT(lying OR hat)

= NOT[NOT(lying) AND NOT(hat)]

= NOT(truth AND NOT(hat))

so A. That man is telling the truth and I won't eat my hat

3. If you don't go to the party and if John goes to the party then I won't go to the party

IF [NOT(you) AND john] THEN NOT(me)

so IF NOT[NOT(you) AND john] THEN NOT[NOT(me)] IF you OR NOT(john) THEN me

So, the answer is:
If you go to the party, or if John does not go to the party, then I will go to the party.


4. Twas brillig, and the slithy toves Did gyre and gimble in the wabe.

Let X = brillig AND gyre AND gimble

Then
NOT(X)
= NOT(brillig AND gyre AND gimble)
= NOT(brillig) OR NOT(gyre) OR NOT(gimble)

so Twasn't brillig, or the slithy toves did not gyre or gimble in the wabe.


Steve used truth tables to work out the negations, making use of NOT(X) is True if and only if X is false

Enumerating the possible combinations for friendly/furry we see that a good pet corresponds to a single row in the truth table. A bad pet is found by negating this.

Friendly Furry Good Pet Bad Pet
0 0 0 1
0 1 0 1
1 0 0 1
1 1 1 0

We can see that this corresponds to 'A bad pet is not friendly or not furry' because it has the same truth table values as the Bad Pet column:

Friendly Furry Not Friendly Not Furry (Not Friendly) OR (Not Furry)
0 0 1 1 1
0 1 1 0 1
1 0 0 1 1
1 1 0 0 0


For the second part, it is confusing because going/not going are opposites. To be clear, the statement is

X: If you don't go to the party and if John goes to the party then I won't go to the party.

This is a little ambiguous because it does not say that I will go to the party in all other cases. I will assume that this is the case, which quickly gives the truth table for me going as:

You go John goes You don't go I go (according to rule X)
0 0 1 1
0 1 1 0
1 0 0 1
1 1 0 1


You go John Goes I go (according to rule NOT(X))
0 0 0
0 1 1
1 0 0
1 1 0

There is only 1 case in which I go now. This is logically the same as If you go to the party, or if John does not go to the party, then I will go to the party.

It is also the same as I won't go if John doesn't go or if you go.



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The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.

NRICH is part of the family of activities in the Millennium Mathematics Project.

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