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Please note that while this is a homework question, I'm simply asking for fact checking and explanations to my solution.

Sentence: If the file is not damaged and the processor is fast, then the printer is slow

My Solution:

Symbolic Translation: $(F'∧ P) → S$

Negation:

  1. $((F'∧ P) → S)'$
  2. $(F ∧ P') → S'$
  3. $((F ∧ P') → (S'))'$
  4. $(F' ∨ P) ∧ S$ --could the negation end here?
  5. $((F')' ∨ P') ∧ S$
  6. $(F ∨ P') ∧ S$

I'm kind of stuck from fixing the errors highlighted by Hunan and Ross. I'm not sure if I should proceed with steps 5 and 6

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  • $\begingroup$ I probably don't understand the problem here. If 0. denotes [(F'∧P)→ S] the negation of 0. comes as 1. ((F'∧ P) → S)'. The negation of 1. comes as logically equivalent to 0. $\endgroup$ – Doug Spoonwood Sep 20 '13 at 14:29
  • $\begingroup$ That makes sense to me, the english translation would then be "If the file is damaged and the processor is not fast, then the printer is not slow. $\endgroup$ – Heidi Sep 20 '13 at 14:52
  • $\begingroup$ No, that wouldn't work as an English translation. You can write (F'∧P) as ∧ 'F P, and (Q→S) as → Q S. If you do this, then [(F'∧P)→ S] can become → ∧ 'F P S. It's negation comes as ' → ∧ 'F P S. I've done this to make it clear you've negated the conditional sentence here. So, the negation of [(F'∧P)→ S] can get read "it is not the case that if the file is not damaged and the processor is fast, then the printer is slow." That comes as truth-functionally equivalent to saying "it is the case that the file is not damaged and the processor is fast, and the printer is not slow." $\endgroup$ – Doug Spoonwood Sep 20 '13 at 15:23
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    $\begingroup$ Which rules did you use to get from 1. ((F'∧ P) → S)' to 2. (F ∧ P') → S'? $\endgroup$ – miracle173 Sep 20 '13 at 15:45
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I check the following statements statement

  1. $((F'∧ P) → S)'$
  2. $(F ∧ P') → S'$

by the truth table generator by Samuel Williams and by John Halleck's Expression Evaluator.

I translate them in a slighly different notation that can be understood by the truth table generator

  1. ~((~F& P) -> S)
  2. (F & ~P) -> ~S

First put ~((~F& P) -> S) in the generator. You get

            !
F  P  S  |  ~  (  (  ~  F  &  P  )  ->  S  )
--------------------------------------------
0  0  0  |  0        1  0  0  0      1  0     
0  0  1  |  0        1  0  0  0      1  1     
0  1  0  |  1        1  0  1  1      0  0     
0  1  1  |  0        1  0  1  1      1  1     
1  0  0  |  0        0  1  0  0      1  0     
1  0  1  |  0        0  1  0  0      1  1   
1  1  0  |  0        0  1  0  1      1  0     
1  1  1  |  0        0  1  0  1      1  1

The result for the expression is ion the column I marked with !

Now do the same with the second expression (F & ~P) > ~S

                               !
F  P  S  |  (  F  &  ~  P  )  ->  ~  S
--------------------------------------
0  0  0  |     0  0  1  0      1  1  0  
0  0  1  |     0  0  1  0      1  0  1  
0  1  0  |     0  0  0  1      1  1  0  
0  1  1  |     0  0  0  1      1  0  1  
1  0  0  |     1  1  1  0      1  1  0  
1  0  1  |     1  1  1  0      0  0  1  
1  1  0  |     1  0  0  1      1  1  0  
1  1  1  |     1  0  0  1      1  0  1  

Th column marked with ! differ so you made an error. Only if both columns match your derivation is right.

For the evaluator we need the following translation

  1. ~((~F& P) > S)
  2. (F & ~P) > ~S

We put (~((~F& P) > S))=((F & ~P) > ~S) in the evaluator and get the result

We find that it is contingent, for example:

  • (p=T s=F ) gives a true evaluation.
  • (p=F s=F ) gives a false evaluation.

If the expression is TRUE the your derivation is ok, if it is contingent or FALSE you made an error.

The truth-table method you can used by calculating by hand. You must not calculate the values for all combinations of $F$, $P$ and $S$ to see that your derivation is false. e.g.for $P=0$, $S=0$, $F=0$ the first expression is $0$ and the second is $1$, so you have an error.

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Step 1 and 2 look the same to me. From step 2 to step 3 is not correct. $A \rightarrow B$ is only false when $A$ is true and $B$ is false. Your transition from step 3 to step 4 is the right type for step 2 to step 3.

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  • $\begingroup$ Hi Ross, thanks for answering. After applying the suggested edits can the negation terminate at step 4? $\endgroup$ – Heidi Sep 20 '13 at 13:17
  • $\begingroup$ The passage from 1 to 2 is still incorrect, as is from 2 to 3 and from 3 to 4. 4 would be a fine end point, were it correct. The rule for negating an implication is $(a \to B)' becomes $A \wedge B'$ $\endgroup$ – Ross Millikan Sep 20 '13 at 17:14
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$\fbox{Original}$ $(\overline{F} \land P) \rightarrow S$ (i've replaced your negation style)

$\fbox{Your attempt to negate it}$

  1. $\overline{((\overline{F} ∧ P)→S)}$ $\color{green}\checkmark$ (hyp)

  2. $\overline{((\overline{F} ∧ P)→\overline{S})}$ $\color{red}\chi$ (doesn't follow from 1)

  3. $\overline{(\overline{F} ∧ P)} → \overline{(\overline{S})}$ $\color{green}\checkmark$ (from 2)

  4. $\overline{(\overline{F} ∧ P)} ∧ S$ $\color{red}\chi$ (doesn't follow from 3)

  5. $(\overline{(\overline{F})} ∨ \overline{P}) ∧ S$ $\color{green}\checkmark$ (from 4)

  6. $(F ∨ \overline{P}) ∧ S$ $\color{green}\checkmark$ (from 5)

So, it fails for those two reasons. You explicitly asked not to give you the solution, so I won't.

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  • $\begingroup$ Hi Hunan, thanks for replying. I edited the negation steps and I'm kind of stuck now. With my most recent edits could the negation end at step 4? $\endgroup$ – Heidi Sep 20 '13 at 13:12
  • $\begingroup$ Np. Now steps (3, 4, 5) don't follow. May I ask why you don't want to get rid of the conditionals and work with disjunction, conjunction, negation, and a bunch of De Morgans? Unless you're explicitly asked to practice with arrows, just transform (A -> B) to (~A v B) and work from there. It should take about three moves to reduce the formula. $\endgroup$ – Hunan Rostomyan Sep 20 '13 at 20:07

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