## Some Logical(?) Arguments

Click on question number to see the analysis

1. If Alice is wrong, then Bill is wrong. If Bill is wrong, then Connie is wrong. Connie is wrong. Therefore, Alice is wrong.

2. If turtles can sing, then artichokes can fly. If artichokes can fly, then turtles can sing and dogs can't play chess. Dogs can play chess if and only if turtles can sing. Therefore, turtles can't sing.

3. If I oversleep, I will miss the bus. If I miss the bus, I'll be late for work unless Sue gives me a ride. If Sue's car is not working, she won't give me a ride. If I am late for work, I'll lose my job unless the boss is away. Sue's car is not working. The boss is not away. Therefore, if I oversleep, I'll lose my job.

### And a selection from Lewis Carroll:

4. No ducks waltz. No officers ever decline to waltz. All my poultry are ducks. Therefore, my poultry are not officers.

5. Everyone who is sane can do Logic. No lunatics are fit to serve on a jury. None of your sons can do Logic. Therefore, none of your sons are fit to serve on a jury.

6. The only articles of food, that my doctor allows me, are such as are not very rich. Nothing that agrees with me is unsuitable for supper. Wedding-cake is always very rich. My doctor allows me all articles of food that are suitable for supper. Therefore, wedding-cake always disagrees with me.

7. Animals, that do not kick, are always unexcitable. Donkeys have no horns. A buffalo can always toss one over a gate. No animals that kick are easy to swallow. No hornless animal can toss one over a gate. All animals are excitable, except buffaloes. Therefore, donkeys are not easy to swallow.

Answer to 1: Let A = "Alice is wrong", B = "Bill is wrong", C = "Connie is wrong". The given statements can now be written symbolically as:
AB,
BC,
and C.
By transitivity of implication, the first two statements give AC. But knowing that C is true and that AC, does not give any information about the truth of A. Thus, this is not a correct argument.

Answer to 2: Let S = "turtles can sing", A = "artichokes can fly" and D = "dogs can play chess". The given statements can now be written symbolically as:

SA,
A(S~D),
DS.
By transitivity, the first two implications give S(S~D). But, since S and D are logically equivalent, this means that S(D~D), which is a contradiction. Since a true statement cannot imply a contradiction, we must have that S is false, i.e., ~S is true - turtles can not sing.

Answer to 3: Let O = "I oversleep", B = "I miss the bus", L = "I'll be late for work", S = "Sue gives me a ride", W = "Sue's car is not working", A = "the boss is away" and J = "I'll lose my job". The given statements can be written symbolically as:

OB,
B(LS),
W~S,
L(JA).
Now, by transitivity, O(LS). But, since W is true and W~S, we have ~S is true. Therefore, OL. Now, since L(JA), we have O(JA), but since ~A is given as true, we can conclude that OJ. This argument is valid.

Answer to 4: Let D(x) = "x is a duck", W(x) = "x waltzes", O(x) = "x is an officer" and P(x) = "x is my poultry". The given statements can be written symbolically (with some equivalent forms) as:

(x)(D(x)~W(x) ~(x)(D(x)W(x)),
(x)(O(x)W(x) ~(x)(O(x)~W(x)),
(x)(P(x)D(x)).
So, for all x, P(x)D(x)~W(x), but by the contrapositive of the second statement we have that for all x, ~W(x)~O(x), so we can conclude that (x)(P(x)~O(x)).

Answer to 5: Let S(x) = "x is sane", L(x) = "x can do Logic", J(x) = "x is fit to serve on a jury" and N(x) = "x is one of your sons". Symbolically we have:

(x)(S(x)L(x)),
(x)(~S(x)~J(x)) ~(x)(~S(x)J(x)),
(x)(N(x)~L(x)) ~(x)(N(x)L(x)).
So, for all x, N(x)~L(x) ~S(x) (by the contrapositive of the first statement)~J(x). Thus, (x)(N(x)~J(x)) ~(x)(N(x)J(x)).

Answer to 6: Let D(x) = "x is allowed by my doctor", R(x) = "x is very rich", A(x) = "x agrees with me", S(x) = "x is suitable for supper", and W(x) = "x is wedding-cake". Symbolically, we have:

(x)(D(x)~R(x)),
(x)(A(x)S(x)),
(x)(W(x)R(x)),
(x)(S(x)D(x)).
Using several contrapositives, we have for all x, W(x)R(x)~D(x)~S(x)~A(x).

Answer to 7: Let K(x) = "x kicks", E(x) = "x is excitable", D(x) = "x is a donkey", H(x) = "x has horns", B(x) = "x is a buffalo", T(x) = "x can toss one over a gate", and S(x) = "x is easy to swallow". Symbolically we have:

(x)(~K(x)~E(x)),
(x)(D(x)~H(x)),
(x)(B(x)T(x)),
(x)(K(x)~S(x)),
(x)(~H(x)~T(x)),
(x)(~B(x)E(x)).
Again, using contrapositives when needed, we have the following chain of implications, for all x, D(x)~H(x)~T(x)~B(x)E(x)K(x)~S(x).