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I am using SD+. Most of the derivation has to be done using it

I am finding this one tricky

I request help or hints to solve it

Derive L => H

1.~L v (~Z v ~U). Assume

2.(U & G) v H Assume

3.Z. Assume

I am thinking the best way is ND.Set up a derivation starting with L and go from there . I will use DS to get H out of the assumptions. But from assumption 2 I need ~( U&G) as ~P to do it, and that is where I am stuck.

I don’t think making so many assumptions helps

Here are some steps.

4.| L. Assume

5.||. ~L Assume ?

6.|| (~Z v ~U ) 5,1 DS ?

7.||~(Z & U) 6 DeM. ?

I feel I am going around in circles.

Note:

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  • The hint is the same already provided in MSE: assume L and derive H from 2nd premise using Disjunction Elim. – Mauro ALLEGRANZA Mar 20 at 8:43
  • That account is closed cause a question l asked was not received well – Eudoxus Mar 20 at 13:54
  • Why are you using incorrect terminology? Unless the problem STAYES there are ASSUMPTIONS you are not to think all.premises are ASSUMPTIONS. There is a distinction between premises that are already KNOWN versus ASSUMPTIONS. CLEARLY an assumption is a statement you are unaware of it's truth value whereas if I stated a fact that is NOT an ASSUMPTION. What textbook are you using? All logic systems do not allow the same rules. You would need to state clearly what set of rules you are using. That is determined by which textbooks are used. Some math proofs use natural deduction rules for instance. – Logikal Mar 20 at 15:05
  • I am not using incorrect terminology. What does STAYES mean ? – Eudoxus Mar 20 at 17:23
  • The Logic Book. Premises are taken as assumptions and we prove the conclusion. For example on page 185 look at the derivation Derive D vB Other than the first 3 assumptions, they assume ~D and derive a contradiction for ~F for the first part of vE – Eudoxus Mar 20 at 17:24
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Well, ~L v (~Z v ~U) is equivalent to L -> (Z -> ~U). So you assume L, then use Z to derive ~U. Now (U & G) v H is equivalent to (~U v ~G) -> H. Because we have ~U we also have ~U v ~G, so this gives us H. We did this under the assumption of L so we get L -> H.

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  • I didn’t see the equivalency. Are you using Exportation to get ( U& G)v H to(~U v~G)->H – Eudoxus Mar 20 at 14:04
  • @Eudoxus A v B is equivalent to ~A -> B and we can use demorgan to get from ~(U&G) -> H to (~U v ~G) -> H. Translate that into SD as you wish. – causative Mar 20 at 14:29

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