Help with Symbolic Logic SD+ Question

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The discussion centers around demonstrating the inconsistency of a specific set of sentences in SD or SD+. The set includes four statements involving logical operations and implications. Participants explore methods to show inconsistency, with one contributor suggesting a substitution and distribution approach to derive a contradiction. Another participant emphasizes that assumptions do not need to be tautologies and provides a step-by-step breakdown of the logical relationships among the premises. They conclude with a method for reaching a contradiction, indicating that further simplification may be possible. The original poster expresses gratitude for the insights and plans to revisit the solutions later.
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Good evening. Would anybody in this room be able to help me with an SD+ question?
My question is as follows:
Show that the following set of sentences is inconsistent in SD or SD+:

{(~C v (E & P)) (triple bar) B, ~E > ~C, ~(P & B) & ~(~P & ~B), B > C}
 
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Marie120 said:
Good evening. Would anybody in this room be able to help me with an SD+ question?
My question is as follows:
Show that the following set of sentences is inconsistent in SD or SD+:

{1.(~C v (E & P)) (triple bar) B, 2.~E > ~C, 3.~(P & B) & ~(~P & ~B), 4.B > C}

Don't know that this is quite what you're looking for, but consider this:

[~C v (E & P)] > C (by substituting for B in 4, from the equivalence given in 1)

[(~C v E) & (~C v P)] > C (By distribution)

That line right there is the same as the argument:

1. ~C v E
2. ~C v P
Therefore, C

which can pretty easily be shown to be invalid. It's a roundabout method, but it should work. I'll leave it to you to write a rigorous proof of this.
 
Loseyourname, you're only working with assumptions. Assumptions do not have to be tautologies. Like if I assume X -> Y, I am not claiming that X, therefore Y, is logically valid for every substitution of X and Y. Such a claim is false but the assumption X -> Y is certainly not inconsistent with itself.

Marie, I don't know about the terms SP or SP+, so if there are special rules I am unaware of then this reply may not be right. But I have worked it through like this:

(your premises)
1. (~C v (E & P)) <--> B
2. ~E --> ~C
3. ~(P & B) & ~(~P & ~B)
4. B --> C

5. B <--> ~P (line 3)
6. (~C v (E & P)) <--> ~P (lines 5, 1)
7. E & P --> ~P
8. ~P v ~(E & P)
9. ~P v ~E v ~P
10. ~P v ~E
11. ~C --> B (line 1)
12. ~C --> C (lines 4, 11)
13. C (line 12)

This is the main part. You can finish it from here. Of course, there may be a simpler way to do it than how I did it, and I didn't formally go into several steps, particularly 5, 7, and 13.
 
Thank you!

Hi

Sorry for my late reply, but I just wanted to say thank you to both Loseyourname and Bartholomew for taking the time to muse over my question. Though I don't have the time right now to apply your solutions to my problem, I definitely will soon.

Have a great New Year! :smile:
 
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