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This is not homework

I do this fun and to expand my knowledge.

I am using the Logic Book

I am having trouble obviously doing SD+ derivations as I post several of them.

I have to show this derivation is valid in SD+ I tried all the extra rules like replacement rules and hasn’t helped.

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Derive I=>~D

1.(F &G )v (H&~I) Assume

  1. I=> ~(F & D). Assume

3.|| I 2 Assume

4.|| ~( F & D ) 2. =>E

5.|| ~~((F & G) v (H & ~I) ) 1 DN

6.|| ~ (~F v ~ G) & ~ (H & ~I). 5DeM

7.|| ~ (~F v ~ G). 6 &E

8.||H & ~I 7. DS

Any help would be appreciated or hints on how to proceed

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  • 1
    It is quite similar to all your other questions... When the conclusion is "A to B", assume A and use the premises to derive B. – Mauro ALLEGRANZA Mar 22 at 10:47
  • Exactly, but it is trying to find the proper rules for each step . I use goal analysis with SD+ ,it works when it is obvious,and drives me nuts otherwise,which is why l post to help me with the steps and rules – Eudoxus Mar 22 at 12:55
  • Just to be clear: the fact that this isn't an assignment from school doesn't make this a question unlike other "homework" questions. It is amazing that you decide to learn this topic by yourself, but you should know that even on SO this type of questions gets downvoted because they are often way too specific to be helpful for others (which is the main point of this network). It'd be more appropriate on dedicated forums (I'm sure reddit has one). – Yechiam Weiss Mar 31 at 17:50
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You want to derive I -> ~D from the premises (F & G) v(H &~I) and I -> (F & D) ? By the way, these are premises not assumptions - you shouldn't write "assume" after them as you are not introducing an assumption scope. Anyway, I -> F & D implies I -> D, and if you also have I -> ~D then we must have that I is false. (Because if I -> D and I -> ~D, then either D or ~D is false, so I implies a falsehood, which must mean I is itself false).

But I need not be false, because the assignment I, F, G, H, D = true satisfies the premises. Therefore we cannot derive I -> ~D from the premises.

To say that again: if we take the premises together with I -> ~D, we reach the conclusion ~I. But from the premises alone, we cannot reach the conclusion ~I. Therefore, the premises cannot imply I -> ~D.

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You have edited your answer to include the correct premises, (F & G) v(H &~I) and I -> ~(F & D) .

Suppose I. Then we obtain ~(F&D). Assume for contradiction F is false. This means F&G is false, so H&~I must be true, which cannot be under the assumption of I. Therefore F is true, which together with ~(F&D) means that D is false. So, this gives us I -> ~D by conditional proof.

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  • That is what it says in the book – Eudoxus Mar 22 at 2:25
  • If it is false then you are stating Bergmann made a mistake. This is the fourth edition. I surely didn’t copy it wrong . You can’t read cause l did put assume. – Eudoxus Mar 22 at 2:39
  • @Eudoxus you did copy it wrong, the book says I -> ~(F & D), not I -> (F & D). You're right that this book uses the strange convention of writing "assume" for the premises. I would prefer to say "premise" for the premises, and reserve "assume" for when you're doing a conditional proof/proof by contradiction. – causative Mar 22 at 2:43
  • @causitive sorry if l insulted you,SD+ is driving me nuts. The question on my mind is how to write the steps. – Eudoxus Mar 22 at 3:31
  • Who downvoted me to -2. Please upvote me. – Eudoxus Mar 22 at 12:26

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