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I'm trying to work out a class exercise and I've got myself stuck. Any help would be appreciated. Thank you.

I am to use use predicate logic reasoning techniques to solve the following problem:

All academics who are computer scientists are programmers or mathematicians. Any logistician is a philosopher. Jack Jones is not a philosopher and he is not a programmer. Prove that if Jack Jones is a logistician, he is not a computer scientist.

Here's what I have done so far.

All academics who are computer scientists are programmers or mathematicians. All (∀) academics who are computer scientists (B) are programmers or mathematicians (C). ∀x[B(x) -> C(x)]

Any logistician is a philosopher. Any (∀) logistician (F) is a philosopher (G). (∃x)[F(x) -> G(x)]

Jack Jones (x) is not a philosopher (H) and he is not a programmer (J). ¬H(x) /\ ¬J(x)

Prove that: if Jack Jones is a logistician, he is not a computer scientist.

  • If Jack Jones is not a programmer then he cannot be a computer scientist.
  • Jack Jones is not a philosopher, however, not all philosophers are logisticians (or not all logisticians are philosophers).
  • Meaning Jack Jones can be a logistician without being a philosopher and a computer scientist.
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  • $\begingroup$ How exactly are you to prove this? You give an informal proof ... is that what you are supposed to do? $\endgroup$
    – Bram28
    Jun 7, 2017 at 13:39

1 Answer 1

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Some of the information is irrelevant, but it can be proven as follows:

  • Cx = x is a computer scientist
  • Px = x is a programmer
  • Lx = x is a logistician
  • Fx = x is a philosopher
  • j = Jack Jones

\begin{array}{l} & \{1\} & 1. & \forall x[Lx \Rightarrow Fx] & \text{ Prem. }\\ & \{2\} & 2. & \neg Fj \land \neg Pj & \text{ Prem. }\\ & \{3\} & 3. & Lj & \text{ Assum. }\\ & \{1\} & 4. & Lj \Rightarrow Fj & \text{ 1 UE }\\ & \{1,3\} & 5. & Fj & \text{ 3,4 MP }\\ & \{6\} & 6. & Cj & \text{ Assum. }\\ & \{1,3,6\} & 7. & Fj \land Cj & \text{ 5,6 $\land$I }\\ & \{1,3,6\} & 8. & Fj & \text{ 7 $\land$E }\\ & \{1,3\} & 9. & Cj \Rightarrow Fj & \text{ 6,8 CP }\\ & \{2\} & 10. & \neg Fj & \text{ 2 $\land$E }\\ & \{1,2,3\} & 11. & \neg Cj & \text{ 9,10 MT }\\ & \{1,2\} & 12. & Lj \Rightarrow \neg Cj & \text{ 3,11 CP }\\ \end{array}

Explanation:

  1. Premise: For every x, if x is a logistician, x is a philosopher.
  2. Premise: Jack Jones is neither a philosopher nor a programmer.
  3. Let's assume that Jack is a logistician.
  4. From our premise on line 1, if Jack is a logistician, he is a philosopher.
  5. It follows from our assumption that he is a philosopher.
  6. Let's assume that Jack is also a computer scientist.
  7. Given these assumptions, Jack would be both philosopher and a computer scientists. (This step may seem redundant, but it's use by logicians to establish a connection between the propositions.)
  8. Jack is a philosopher. (Because of line 7, this assertion is now shown to depend on line 1, 3 and 6, whereas before it depended only on line 6. Again, it's used to establish a logical connection.)
  9. If Jack is a logistician, he is a philosopher. (Because line 7 is shown to depend on line 3, line 7 is considered to be a logical consequence of 3, establishing the implication. The assumption on 3 is discharged.)
  10. According to the premise on 2, Jack is not a philosopher;
  11. Therefore Jack cannot be a logistician. (This is the application of Modus Tollens, together with the implication on line 9. If Jack were a logistician, line 9 says that he would also be a philosopher, contradicting the premise.)
  12. Based on the assumption on line 3, the conclusion is established: If Jack is a logistician, he is not a computer scientist. (The assumption on 3 is also discharged.)
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  • $\begingroup$ Thank you for your answer. Is there any chance you could explain it step by step please? $\endgroup$
    – Mcclaine
    Jun 11, 2017 at 15:37
  • $\begingroup$ @Mcclaine. I edited my answer to explain the steps. $\endgroup$
    – User4407
    Jun 11, 2017 at 16:03

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