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  1. This is not homework .

  2. l do this for fun. I am 58.

  3. I graduated Concordia University in 1993

  4. Problem comes from the Logic Book by Bergman pg 217 (17)c

I want to see if my SL is correct.

If civil disobedience is moral,then not all resistance to the law is morally prohibited,although our legal code is correct if all resistance to the law is morally prohibited .But civil disobedience is moral iff either civil disobedience is moral or our legal code is correct.Our judges have acted well only if all resistance to the law is morally prohibited.So our judges haven’t acted well.

Let A=“civil disobedience is moral”

B=“All resistance to the law is morally prohibited”

C=“our legal code is correct “

D=“judges act well”
  1. (A->~B)&(D->B)&(A=A v D)

  2. D->A

———————————————————-————

~D

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  • "our legal code is correct if all resistance to the law is morally prohibited" must be "if B, then C" – Mauro ALLEGRANZA Mar 11 at 14:14
  • Thanks for the correction – Eudoxus Mar 11 at 14:16
  • "Our judges have acted well only if all resistance to the law is morally prohibited" must be "if D, then B" – Mauro ALLEGRANZA Mar 11 at 14:16
  • Thanks again for the correction – Eudoxus Mar 11 at 14:17
  • "civil disobedience is moral iff either civil disobedience is moral or our legal code is correct" must be "A iff (A or C)", that sound quite weird... – Mauro ALLEGRANZA Mar 11 at 14:18
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If we tidy up the text a little and divide it up into separate premises it becomes:

  1. If civil disobedience is moral (A) then not all resistance to the law is morally prohibited (¬B)
  2. Our legal code is correct (C) if all resistance to the law is morally prohibited (B)
  3. Civil disobedience is moral (A) iff either civil disobedience is moral (A) or our legal code is correct (C)
  4. Our judges have acted well (D) only if all resistance to the law is morally prohibited (B)
  5. Conclusion: Our judges haven’t acted well (¬D)
1. A → ¬B
2. B → C
3. A ↔ A v C  (simplifies to C → A since A ↔ A is a tautology)
4. D → B
5. ¬D

This is valid in classical logic. You could demonstrate it using a truth table of ((A → ¬B) ∧ (B → C) ∧ (C → A ) ∧ (D → B)) → ¬D. Or if you prefer a natural deduction proof, 3 together with 1 gives you C → ¬B by hypothetical syllogism, while the contraposition of 2 gives ¬C → ¬B, and since C v ¬C (by LEM), you can derive ¬B by disjunction elimination, and thence ¬D by modus tollens.

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  • I did not think of setting them up as separate premises . I haven’t got into SD+ yet. Thanks – Eudoxus Mar 11 at 18:43

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