Help solving symbolic logic proof?

Question

P v (Q v R) ⊢ Q v (P v R)

Answer 1

P v (Q v R) /- Q v (P v R)



(1) 1. P v (Q v R) Premise

(2) 2. P Assumption

(2) 3. P v R 2 vI

(2) 4. Q v (P v R) 3 vI

(5) 5. Q v R Assumption

(6) 6. Q Assumption

(6) 7. Q v (P v R) 6 vI

(8) 8. R Assumption

(8) 9. P v R 8 vI

(8) 10. Q v (P v R) 9 vI

(5) 11. Q v (P v R) 5,6,7,8,10 vE

(1) 12. Q v (P v R) 1,2,4,5,11 vE



It doesn't make sense to speak of THE rules of inference; there are many sets of such rules. Given that the asker wants to use vE (see his other question), it is quite clear that my proof is ideal here, as it is constructed within a system that uses that rule. Neither of the other proofs that have been offered here is constructed in an appropriate system for this particular asker.



Of course, you could design a system that treats P > P as an axiom schema, but it would be hard to ensure that it were non-redundant. The most popular axiom set is Frege's, in which P > P can easily be proved as a theorem in five lines.



Some would argue that systems that use vE are preferable to Copi's and Hurley's because (i) they contain fewer rules, and (ii) each of their rules is essential (removing it from the system renders the system incomplete). It is a virtue of the system I use that it makes it quite clear, for any given line, on what the formula on that line depends; in Copi's system, for example, such dependencies are not immediately clear.



I grant that in the case of propositional logic, the last point isn't all that important, but it makes a significant difference in predicate logic. For example, assuming we don't have recourse to quantifier-switch rules, any proof of the theoremhood of, say, (Vx)(Fx > (Ey)(Gy & Rxy)) <> (Vx)(Ey)(Fx > (Gy & Rxy)), will be decidedly unclear unless we have some way of displaying dependencies.



It's also worth pointing out that Copi's system does not permit the derivation of formulas as theorems unless we add to it some such rule as conditional proof or RAA (basically, a rule that allows us to discharge assumptions), but that just adds to the (already needlessly long) list of rules.



As a final point, if we're getting into questions of fundamentality, CP is far less controversial than material implication, which, insofar as it is used to introduce the material conditional, is CP's equivalent in such systems as Copi's. No one objects to CP, whereas plenty of people take issue with material implication. A strong case can be made for supposing it necessary for one's grasping the meaning of 'if' that one be disposed to reason in accordance with CP, but it would be absurd to regard being disposed to employ material implication as in any way necessary for such understanding. Indeed, many take the so-called paradoxes of material implication to demonstrate that 'if' and the material conditional are not even equivalent.

Answer 2

Logic Proof Calculator

Answer 3

Q=P=6

Answer 4

Logic Proof Solver

Answer 5

Here's a direct proof that doesn't assume disjunction is commutative, or associative, or anything. You oughtn't to need anything more fundamental than this---though I suppose there are systems of propositional logic so minimalist that it's still possible to nitpick.



01. P → P [Axiom]

02. P → (P ∨ R) [02, disjunction introduction on right]

03. P → (Q ∨ (P ∨ R)) [03, disjunction introduction on left]



04. Q → Q [Axiom]

05. Q → Q ∨ (P ∨ R) [04, disjunction introduction on right]



06. R → R [Axiom]

07. R → (P ∨ R) [06, disjunction introduction on left]

08. R → (Q ∨ (P ∨ R)) [07, disjunction introduction on left]



09. (Q ∨ R) → (Q ∨ (P ∨ R)) [05, 08, disjunction elimination]

10. (P ∨ (Q ∨ R)) → (Q ∨ (P ∨ R)) [03, 09, disjunction elimination]



11. P ∨ (Q ∨ R) [Given]

12. Q ∨ (P ∨ R) [10, 11, modus ponens]

Answer 6

From here

P v (Q v R)



The law of associativity allows you to do this..

(P v Q) v R



Then the law of commutativity allows you to do this...

(Q v P) v R



And the law of associativity again allows you to do this...

Q v (P v R



====== Update ==

Question for you...

Other people are treating this question differently than I am. Did you want this propositions proven using the rules of inference that you were taught? Or do you want to prove the rules of inferences themselves from first principles? Im not taking the "from axiom" approach since you didnt ask anyone to do so.



Proving complicated expressions such as yours from axioms is an unnecessary over-complication. Use the rules of inference. If you need the rules of inference themselves proved *then* use axioms. There is no intelligent reason to go from axiom to some complicated expression directly.



Ian takes the long route from axioms. Not to mention that some schools of thought do not consider P⊃P as an axiom, and under these schools of thought he is just using another rule of inference. Anisha, on the other hand, is relying on the assumption that you can reason from assumption... she has presumed that conditional proofs are allowable as a means to an end. Her hypothetical P leads to the conclusion that P⊃P∨X, some other expression. From what I can tell. I perceive that as yet another rule of inference. I am guilty of the same, I rely on the rules of replacement (but not the rules of implication nor reasoning from hypotheses). The only assumptions I have made (and we all made them) was the rules of replacement... which can just as easily themselves be proved from axioms. But guess what? Even the axioms themselves are unproven assumptions.

Answer 7

This Site Might Help You.



RE:

Help solving symbolic logic proof?

P v (Q v R) ⊢ Q v (P v R)

Answer 8

What you want to prove is that A v B and A v C IMPLIED A v (B v C). So construct a truth table Columns: 1....2.....3........4..........5......... A....B....C.....A v B....A v C....(AvB) and (AvC).......B v C.....A v (B v C) T....T.....T........T...........T-----... T....T.....F........T...........T-----... T....F.....T........T...........T-----... T....F.....F........T...........T-----... F....T.....T........T...........T-----... F....T.....F........T...........F-----... F....F.....T........F..........T------... F....F.....F........F..........F------... To see that column 6 implies column 8, you can either construct a new column for: [(A v B) and (A v C)] => [A v (B v C)], or you can directly realize that whenever column 6 is true, column 8 is also true. (EDIT: I don't know why but when I submit my answer, some of the T's and F's are not appearing, as you have noticed. Anyway you can just complete the truth table as an exercise.)