# Thread: logic and sets help!!

1. ## logic and sets help!!

can some plz help me figure this out:

Consider the following collection of statement.:-
If I wash thent.

iii. How many rows would be required by a truth table to show that the
statements are inconsistent?

2. Hello, srk619!

Consider the following collection of statements:

(1) If I wash the dishes, I break something.
(2) If I break something, I get into trouble.
(3) If I do not wash the dishes, I am accused of being lazy.
(4) If I am accused of being lazy, I get into trouble.
(5) If I get into trouble, I will sulk.
(6) I will not sulk.

i. Choose symbols to represent the sentences in the above argument,
and hence express the statements in terms of propositional calculus.
. . $\begin{array}{ccc}w & = & \text{I wash the dishes} \\ b &=& \text{I break something} \\ t &=& \text{I get into trouble} \\ l &=& \text{I'm accused of being lazy} \\ s &=& \text{I will sulk} \end{array}$

And we have: . $\begin{array}{cc}(1) & w \to b \\ (2) & b \to t \\ (3) & \sim w \to l \\ (4) & l \to t \\ (5) & t \to s \\ (6) & \sim s \end{array}$

ii. Construct a formal proof showing that the statements are inconsistent.
Working on it . . .

iii. How many rows would be required by a truth table
to show that the statements are inconsistent?

There are five statements: . $w, b, t, l, s$

The truth table will have: . $2^5 = 32$ rows.

3. ## formal proof

i dont know how to do the formal proof can you help me

4. Originally Posted by srk619
i dont know how to do the formal proof can you help me
using Soroban's numbering, let statements 1 through 5 be premises, and statement 6 the conclusion. show that such an argument is invalid.

(note that in either case, whether you wash the dishes or not, you will get into trouble. statement 5 says if you get into trouble you will sulk, but statement 6 says you won't. obviously this is inconsistent. now find the proof. statement 5 and 6 cannot both be true if statements 1 through 4 are true.)

5. Hello, srk619!

. . $\begin{array}{ccc}w & = & \text{I wash the dishes} \\ b &=& \text{I break something} \\ t &=& \text{I get into trouble} \\ l &=& \text{I'm accused of being lazy} \\ s &=& \text{I will sulk} \end{array}$

And we have: . $\begin{array}{cc}(1) & w \to b \\ (2) & b \to t \\ (3) & \sim w \to l \\ (4) & l \to t \\ (5) & t \to s \\ (6) & \sim s \end{array}$

ii. Construct a formal proof showing that the statements are inconsistent.
There must be hundreds of ways to approach this proof . . . Here's mine.

Let $CP(\;)$ = "contrapositive of"
. . For example: . $CP(3)$ means "contrapositive of statement (3)".

. . $\begin{array}{cccc}& \text{Statement} & & \text{Reason} \\ \hline \\[-4mm]
(a) & \sim s \to \sim t & & CP(5) \\
(b) &\sim t \to \sim l & & CP(4) \\
(c) & \sim l \to w & & CP(3) \\
(d) & w \to b & & (1) \\
(e) & b \to t & & (2) \\
-- & ---- \\
(f) & \sim s \to t & & \text{Syllogism} \\\end{array}$

But this contradicts statement $(a)\!:\;\;\sim s \to \sim t$

Therefore, the statements are inconsistent.

6. I've got the exact same question to answer

But i still don't really understand how you got to that conclusion.

I tried to write the proof myself but ended up with:

W = I Wash The Dishes, B = I Break Something, T = I Get In Trouble, L = I am Accused Of Being Lazy,
S = I Will Sulk

H1 = W => B [If I wash the dishes, I break something]
H2 = B => T [If I break something, I get in trouble]
H3 = ¬W => L [If I do not wash the dishes I am accused of being lazy]
H4 = L => T [If I am accused of being lazy I get in trouble]
H5 = T => S [If I get in trouble I will sulk]
H6 = ¬S [I will not sulk]

Assertion | Justification
1. ¬S | H6
2. T => S | H5
3. ¬T | 1, 2 Modus Tollens
4. L => T | H4
5. ¬L | 3, 4 Modus Tollens
6. B => T | H2
7. ¬B | 3, 6 Modus Tollens
8. ¬W => L | H3
9. W | 5, 8 Modus Tollens
10. W => B | H1

But that doesn't seem to prove anything, plus i think part of it is wrong

Could anyone help by giving a step by step walkthrough of what you are supposed to do?

Thank you.

7. Originally Posted by Soroban
Hello, srk619!

. . $\begin{array}{ccc}w & = & \text{I wash the dishes} \\ b &=& \text{I break something} \\ t &=& \text{I get into trouble} \\ l &=& \text{I'm accused of being lazy} \\ s &=& \text{I will sulk} \end{array}$

And we have: . $\begin{array}{cc}(1) & w \to b \\ (2) & b \to t \\ (3) & \sim w \to l \\ (4) & l \to t \\ (5) & t \to s \\ (6) & \sim s \end{array}$

There must be hundreds of ways to approach this proof . . . Here's mine.

Let $CP(\" alt="CP(\" /> = "contrapositive of"
. . For example: . $CP(3)$ means "contrapositive of statement (3)".

. . $\begin{array}{cccc}& \text{Statement} & & \text{Reason} \\ \hline \\[-4mm]
(a) & \sim s \to \sim t & & CP(5) \\
(b) &\sim t \to \sim l & & CP(4) \\
(c) & \sim l \to w & & CP(3) \\
(d) & w \to b & & (1) \\
(e) & b \to t & & (2) \\
-- & ---- \\
(f) & \sim s \to t & & \text{Syllogism} \\\end{array}$

But this contradicts statement $(a)\!:\;\;\sim s \to \sim t$

Therefore, the statements are inconsistent.

Forgive me, as Im new to this area of maths.

Im sorry but Im trying to understand your proof, or see alternatives, but Im struggling to see how you've made this case...mainly the area of syllogism? Now if its hypothetical syllogism, shouldn't (f) be w --> t

I struggle to see how ~s --> t comes from statment 1 and 2.

Maybe you can point me in the direction of another approach, if syllogism is a complicated kettle of fish

8. Are there statements not omitted between (e) and (f) that follow the rules of inference, is this the reason why? I cant see why statement 1 and 2, are present in the tableau if they are not worked on?

9. never mind...I eventually got there.

Contra positive is new to me though. Do you have any direct online reading material, or recommended math web links?

Contra positive by the way, that isn't de Morgans Law is it?