Professor Kate Poirier, Spring 2017

Category: Test #1 Solutions

test solution #6b

a=superman is able to prevent evil

b= superman is impotent

c=superman is willing to prevent evil

d=superman is malevolent

e= superman exsists

f= superman prevents evil

 

argument form=

(a∧c)→f

¬a→b

¬c→d

¬f

e→¬b∧¬d

___________________

:.¬e

 

Argument =

(we have to prove ¬e)

 

  1.  (a∧c)→f    =”p”→f  =add  !f  =  apply modus tollens   (!f∧(p→f))→!p = !p = !(a∧c)
  2.  ¬ (a∧c)  =  !a V !c

¬a→b

¬c→d

 

e→¬(b∧d)

 

*i am lost after this

i know this is supposed to look like this

and pg 73 of the book has it but i dont really understand how the substituting works

 

this website explains more about building arguments:

http://sites.millersville.edu/bikenaga/math-proof/rules-of-inference/rules-of-inference.html

 

 

 

 

 

 

 

 

 

 

 

 

 

Test solution #3

3a

T(a,b) = Student a likes TV show b where the domain for a is the students in the class and the domain for b is the shows currently airing on TV

Translate:

∀s∃t∃u(T(s,u)→¬T(t,u))

T(s,u) = all students s like a TV show u

or everybody likes a TV show

¬T(t,u) =  it is not the case that a student t likes the TV show u

its not true that one student likes the TV show

(T(s,u)→¬T(t,u))

if everybody likes a TV show, then its not true that one student likes the TV show,

* can this be translated into :

if everybody likes a TV show, then there is not one student who doesnt like the TV show?

3b

use quantifiers and variable to translate the following prop

there is a TV show that everybody likes.

on the test i put this:

∀x∃y (T(x,y))

*should it be

∃y∀x (T(x,y))?

 

Test 1 solution #4

4a. Prove that if x is a rational number and x ≠ 0 then 1/x is a rational number

If x is a rational number ≠ 0, then it can be written as a/b, where a and b are integers and ≠ 0. Then 1/x is equal to 1/(a/b) which equals b/a. Therefore if x is a rational number, 1/x is also rational.

4b. Prove that if x is an irrational number and x≠ 0 then 1/x is an irrational number.

Assume that x is irrational and 1/x is rational. Then 1/x can be written as a/b where a and b are integers ≠ 0. So x is equal to 1/(a/b) which equals b/a, b and a are integers. Therefore x is rational. This contradicts the assumption that x is irrational. Therefore it is true that if x is irrational, 1/x is also irrational.

Test #1 Solutions Question 6a

6a. Determine the argument form of the above argument.

Notation: W(x): x is willing to prevent evil

A(x): x is able to prevent evil

P(x): x prevents evil

I(x): x is impotent

M(x): x is malevolent

E(x): x exists.

S stands for Superman.

Let’s write the givens:

1: (A(S) ∧ W(S)) → P(S)

2. ¬A(S) → I(S)

3. ¬W(S) → M(S)

4. ¬P(S)

5. E(S) → (¬M(S) ∧ ¬I(S))

6. ¬A(S) ∨ ¬W(S)

7. M(S) ∨ I(S)

8. ¬(¬M(S) ∧ ¬I(S))

9. ¬E(S)

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