NOTty logic
Have a go at being mathematically negative, by negating these statements.
Problem
The negation NOT$(P)$ of a statement is true if and only if the statement $P$ is false. A well-constructed negation uses positive language, avoiding the use of the word NOT.
Here are two statements, each with four suggested negations. Which of these are the correct negations and why?
1. A good pet is friendly and furry
A. A good pet is unfriendly and unfurry
B. A bad pet is friendly and furry
C. A good pet is unfriendly or unfurry
D. A bad pet is unfriendly or unfurry
2. That man is lying or I'll eat my hat
A. That man is telling the truth and I won't eat my hat
B. Either that man is telling the truth or I won't eat my hat
C. I won't eat my hat or that man is telling the truth
D. That man is telling the truth and I don't have a hat
Have a go at negating this sentence:
If you don't go to the party and if John goes to the party then I won't go to the party
Finally, try to negate this sentence taken from Lewis Carrol's Jabberwocky from Through the Looking-Glass and What Alice Found There, 1872
Twas brillig, and the slithy toves
Did gyre and gimble in the wabe.
This question is based on an exercise from A Mathematical Bridge (2nd ed), by Stephen Hewson. Published by World Scientific.
Getting Started
Look at the variables in each statement which can either, individually, be true or false.
Then draw a truth table for the statements.
The NOT of the statement will be created by switching 1s to 0s, and vice versa, in the appropriate part of the truth table.
Student Solutions
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.
Teachers' Resources
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