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i am new in logic course, i saw an question in my book. it says that show the following argument is valid by using natural derivation rules.

if [if (if P then Q) then (if P then R)] then [if P then (if Q then R)]

However, it is very advanced for me and i stuck in it.I could not derive it. Can you help me? if you can derive it step by step i can apply it in other problems.

  • What textbook are you using and which rules are you allowed to use. There can be different rules allowed such as natural deduction rules or another set of rules. Does your text list the rules on the inside cover of the text? – Logikal Aug 21 '20 at 20:42
  • @Logikal all rules can be used from MP to BC such as adjuction,addition mtp, biconditonal-conditional,simplification etc. – martin scott Aug 21 '20 at 20:44
  • @Logikal conditional proof,indirect proof, separation etc can be used – martin scott Aug 21 '20 at 20:45
  • Are you sure you represented the problem correctly? You should put the problem in a clear distinct form. You should identify each premise and separate the conclusion from the premises. You wrote everything as one big premise. The form should appear like an addition problem: premise one line one, premise two line two and the conclusion on the third line. In this way you can visually see the form of the argument as a whole. Is this a math class you are taking or a philosophy class? – Logikal Aug 21 '20 at 22:02
  • @logical it is a proof of a theorem, there is no any premise and philosophy class – martin scott Aug 22 '20 at 2:32
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If I understand you correctly, you are being asked to prove that the sentence

[(P → Q) → (P → R)] → [P → (Q → R)]

is a tautology, or a valid sentence of propositional logic. I'm also assuming your conditional should be represented as the material conditional, since this is the one you are most likely using in an introductory course on logic. I've replaced your "if/then" with →.

A good rule of thumb when being asked to prove a conditional is to work backwards from the conclusion. Treat the antecedent of the conditional as an assumption, then prove the consequent part, so that your final step is discharging that assumption by the rule of conditional proof. So you want to assume (P → Q) → (P → R) and prove P → (Q → R). Since P → (Q → R) is also a conditional, do the same thing with that: assume P, prove Q → R then discharge the assumption using conditional proof. Since Q → R is also a conditional, do the same thing again: assume Q and prove R. Put together it looks like this:

1. (P → Q) → (P → R)                        Ass. 
2. P                                        Ass. 
3. Q                                        Ass.
4. P → Q                                    3, Impl. 
5. P → R                                    1,4 MP
6. R                                        5,2 MP
7. Q → R                                    6,3 CP 
8. P → (Q → R)                              7,2 CP
9. [(P → Q) → (P → R)] -> [P → (Q → R)]     8,1 CP

At step 4, I've assumed that you have a rule for material implication that allows you to go from Q to P → Q. If you are not allowed this rule, then you will need an extra step to prove that.

  • sir thank you very much but my book does not allow to go from Qto if P then Q. if you have time can you prove it.it would be guideline for me.Thank you again – martin scott Aug 22 '20 at 2:28
  • by the way the rule which is written CP is not allowed. and i think that tbe proof must be longer than you did.if you have free time can you write it more simple for me – martin scott Aug 22 '20 at 2:41
  • To get P → Q from Q, you can use disjunction introduction to go from Q to ¬P v Q, and this is equivalent to P → Q. If you don't have that rule either, you can just assume P and discharge the assumption to get P → Q. Of course, you already have the assumption P at step 2, so you could also just discharge on that. – Bumble Aug 22 '20 at 2:44
  • The rule CP is just conditional proof. I cannot imagine why you would not be allowed that, since it is the standard rule of conditional introduction. What rule of conditional introduction do you have? – Bumble Aug 22 '20 at 2:45

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