Philosophy Now Forum

Hey all,

Can anyone help prove this using the following rules? I was a philosophy major with a concentration in logic, but I graduated a long time ago. Lately, I've been trying to shake the rust off by working through Copi's Introduction to logic 5ed, but I've been working on this one in particular for way too long now.

Thanks for any help!

1. ~J v (~J&K)
2. J -> L. //...(L&J) <-> J

Available Rules:
I. Rules of Inference (10)


1. Modus Ponens (M.P.):
P⊃Q
P
∴ Q

2. Modus Tollens (M.T.):
P⊃Q
∼Q
∴ ∼P

3. Disjunctive Syllogism (D.S.):
P˅Q
∼P
∴ Q

4. Constructive Dilemma (C.D.):
(P⊃Q) ˅ (R⊃S)
P˅R
∴ Q˅S

5. Destructive Dilemma (D.D.):
(P⊃Q) ˅ (R⊃S)
∼Q ˅ ∼S
∴ ∼P ˅ ∼R


6. Hypothetical Syllogism (H.S.):
P⊃Q
Q⊃R
∴ P⊃R

7. Simplification (simp.):
P•Q
∴ P


8. Conjunction (conj.):
P
Q
∴ P•Q


9. Addition (add.)
P
∴ P˅Q


10. Absorption (abs.):
P⊃Q
∴ P ⊃ (P•Q)


II. Rules of Replacement (11)


11. De Morgan’s Theorems (DeM.):
∼ (P˅Q) ≡ (∼P • ∼Q)
∼ (P•Q) ≡ (∼P ˅ ∼Q)

12. Commutation (comm.):
(P˅Q) ≡ (Q˅P)
(P•Q) ≡ (Q•P)

13. Association (assoc.):
[P ˅ (Q˅R)] ≡ [(P˅Q) ˅ R]
[P • (Q•R)] ≡ [(P•Q) • R]

14. Distribution (dist.):
[P • (Q˅R)] ≡ [(P•Q) ˅ (P•R)]
[P ˅ (Q•R)] ≡ [(PvQ) • (P˅R)]

15. Double Negation (D.N.): P ≡ ∼∼P

16. Transposition (trans.): (P⊃Q) ≡ (∼Q ⊃ ∼P)

17. Material Implication (imp.): (∼P ˅ Q) ≡ (P⊃Q)

18. Material Equivalence (equiv.):
(P≡Q) ≡ [(P⊃Q) • (Q⊃P)]
(P≡Q) ≡ [(P•Q) ˅ (∼P • ∼Q)]

19. Exportation (exp.): [(P•Q) ⊃ R] ≡ [P ⊃ (Q⊃R)]

20. Tautology (taut.):
P ≡ (P˅P)
P ≡ (P•P)

21. Negated Conditional (N.C.) ~(P⊃Q) ≡ P • ~Q

NSRoss wrote: ↑Sat Oct 05, 2019 9:10 pm Hey all,

Can anyone help prove this using the following rules? I was a philosophy major with a concentration in logic, but I graduated a long time ago. Lately, I've been trying to shake the rust off by working through Copi's Introduction to logic 5ed, but I've been working on this one in particular for way too long now.

Thanks for any help!

1. ~J v (~J&K)
2. J -> L. //...(L&J) <-> J

Available Rules:
I. Rules of Inference (10)


1. Modus Ponens (M.P.):
P⊃Q
P
∴ Q

2. Modus Tollens (M.T.):
P⊃Q
∼Q
∴ ∼P

3. Disjunctive Syllogism (D.S.):
P˅Q
∼P
∴ Q

4. Constructive Dilemma (C.D.):
(P⊃Q) ˅ (R⊃S)
P˅R
∴ Q˅S

5. Destructive Dilemma (D.D.):
(P⊃Q) ˅ (R⊃S)
∼Q ˅ ∼S
∴ ∼P ˅ ∼R


6. Hypothetical Syllogism (H.S.):
P⊃Q
Q⊃R
∴ P⊃R

7. Simplification (simp.):
P•Q
∴ P


8. Conjunction (conj.):
P
Q
∴ P•Q


9. Addition (add.)
P
∴ P˅Q


10. Absorption (abs.):
P⊃Q
∴ P ⊃ (P•Q)


II. Rules of Replacement (11)


11. De Morgan’s Theorems (DeM.):
∼ (P˅Q) ≡ (∼P • ∼Q)
∼ (P•Q) ≡ (∼P ˅ ∼Q)

12. Commutation (comm.):
(P˅Q) ≡ (Q˅P)
(P•Q) ≡ (Q•P)

13. Association (assoc.):
[P ˅ (Q˅R)] ≡ [(P˅Q) ˅ R]
[P • (Q•R)] ≡ [(P•Q) • R]

14. Distribution (dist.):
[P • (Q˅R)] ≡ [(P•Q) ˅ (P•R)]
[P ˅ (Q•R)] ≡ [(PvQ) • (P˅R)]

15. Double Negation (D.N.): P ≡ ∼∼P

16. Transposition (trans.): (P⊃Q) ≡ (∼Q ⊃ ∼P)

17. Material Implication (imp.): (∼P ˅ Q) ≡ (P⊃Q)

18. Material Equivalence (equiv.):
(P≡Q) ≡ [(P⊃Q) • (Q⊃P)]
(P≡Q) ≡ [(P•Q) ˅ (∼P • ∼Q)]

19. Exportation (exp.): [(P•Q) ⊃ R] ≡ [P ⊃ (Q⊃R)]

20. Tautology (taut.):
P ≡ (P˅P)
P ≡ (P•P)

21. Negated Conditional (N.C.) ~(P⊃Q) ≡ P • ~Q

PM Avicenna or Wtf if they are around.

NSRoss wrote: ↑Sat Oct 05, 2019 9:10 pm Hey all,

Can anyone help prove this using the following rules? I was a philosophy major with a concentration in logic, but I graduated a long time ago. Lately, I've been trying to shake the rust off by working through Copi's Introduction to logic 5ed, but I've been working on this one in particular for way too long now.

Thanks for any help!

1. ~J v (~J&K)
2. J -> L. //...(L&J) <-> J

I can try at least! However, the symbols “//...” are not clear to me. For a couple of reasons, I have taken it to mean that it could be part of a larger statement. However, for the purposes of my contribution, this is how I will take it:

1. ~J v (~J&K)
2. J -> L.
Therefore, (L&J) <-> J

The conclusion is a bi-conditional. So, by rule 18, it is equivalent to:

(L&J)↔J ≡ ((L&J)→J) & (J → (L&J))

Since, we have a conjunction to prove, we can prove each conjunct separately, I.e we will prove:

(i) ((L&J)→J)), and then
(ii) (J → (L&J)), before conjoining them with rule 8 in the final step of the proof.

The first conjunct, i.e. ((L&J)→J)), is in fact a tautology, i.e. it does not depend on any premises. This is easily prove, by assuming it’s negation and showing a contradiction entails.

1. ~((L&J)→J)) Assumption
2. (L&J) & ~J applying rule 21 to statement (1)
3. L & J applying rule 7 to (2)
4. J applying rule 7 to (3)
5. ~ J applying rule 7 to (2)
6. contradiction in statement (4) and (5)


The second conjunct, is even more easily proved as it is a direct application of the absorption rule (rule 10) you provided, to premise 2 you gave!

1. J → L
2. (J → (L&J)) Absorption rule (rule 10) to statement (1)

Since we have proved both conjuncts, we can conjoin them in a last step, using rule 8. And that’s it!!

Now, one thing that makes me think that this could be part of a larger statement is the fact that I had no need of the premise 1 you provided. The proof rests only on the truth of premise 2. In fact the proof is of the following statement:

1. J -> L.
Therefore, (L&J) <-> J

The premise ~J v (~J&K) is redundant in my proof. This is not a problem though. Redundant premises does not invalidate valid arguments!
I hope this helps.