Solving a>b, b>-c, and a>-c: HSTC

  • Thread starter drema9
  • Start date
In summary, to solve the problem of finding -a v -c, we can use the rules of natural deduction to deduce statements from the given premises. By using simplification and hypothetical syllogism, we can show that a logically equivalent statement to -a v -c is ~e v ~a. By using modus tollens and double negation, we can deduce ~e->~a, thus proving that -a v -c is true.
  • #1
drema9
4
0
the question is.
1. (a>b). (b>c)
2. c>-d
3. b>e
4. -d>f
5. -e v -f

We have to find -a v -c

i can get so far then bam nothing! any help would be great thank you.

6. a>b 1 simp
7. b>-c 1,6, mp
8. a>-c h.s.
 
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  • #2
I'm confused by the first line. Is it A[tex]\rightarrow[/tex]B and B[tex]\rightarrow[/tex]C ?

Anyways, you're close to a solution in 8. Do you know a relation between the implication and or operators? In other words, do you know a statement using implication that is logically equivalent to the statement -a v -c ?

The problem is I don't follow your deduction of line 7, how do you get b>-c from a>b and b>c?

If you can deduce -a v -c you're practically done.
 
Last edited:
  • #3
thks

i deduced line seven (b>-c)
taking line one which is ( a>b) and (b>-c)
and line six (a>b).

I don't know how to get the -a v-c.
 
  • #4
I see... In your original post you have b>c. Your deduction is correct.
 
  • #5
cool

cool,
any idea what a next step would be? this one is driving me nuts
 
  • #6
I think you only need to find -A and you can make an Add to get -A v -C.
 
  • #7
drema9 said:
the question is.
1. (a>b). (b>c)

I don't understand the (.) operator, What does mean?
 
  • #8
drema9 said:
the question is.
1. (a>b). (b>c)
2. c>-d
3. b>e
4. -d>f
5. -e v -f

We have to find -a v -c

i can get so far then bam nothing! any help would be great thank you.

6. a>b 1 simp
7. b>-c 1,6, mp
8. a>-c h.s.

Hey you only need one step. If you reach to step 8 then you only need to make an implication of 8.

Implication
P[tex]\rightarrow[/tex]Q:: -P v Q
 
  • #9
ok, i am still stuck i can not get -a then i can wedge in the last part.
 
  • #10
drema9 said:
ok, i am still stuck i can not get -a then i can wedge in the last part.

Why?
If you have a>-c by implication you have -a v -c
 
  • #11
some remarks, if you intended already to deduce from (a->b)&(b->c) by mp b->c why didn't you use simplification on it?
anyway here's one proof:
1. (a->b). (b->c)
2. c->~d
3. b->e
4. ~d->f
5. ~e v ~f
6. a->b 1,simp
7. a->e 3,6 hypothetical syllogism.
8. f->~e 5,material conditional.
9. ~d->~e 4,8, hypo syllogsim
10. c->~e 2,9 h.s
11. ~~e->~c 10, modus tollens.
12. e->~c 11, double negation.
13. ~e->~a 7,modus tollens.
now you can take it yourself.

p.s in questions in logic in the future take care on describing which rules of natural deduction you can use and which you cannot!
 

1. How do you solve a>b, b>-c, and a>-c?

To solve this problem, you need to apply the transitive property of inequality. First, subtract b from both sides of the second inequality to get a>b and a>-c-b. Then, combine the two inequalities to get a>-c. This shows that a is greater than -c, satisfying all three conditions.

2. What is the transitive property of inequality?

The transitive property of inequality states that if A>B and B>C, then A>C. In other words, if one quantity is greater than another and that second quantity is greater than a third quantity, then the first quantity must also be greater than the third quantity.

3. Can you provide an example of how to solve a>b, b>-c, and a>-c?

Sure, let's say a=5, b=3, and c=-2. First, we have 5>3, 3>-2, and 5>-2. Subtracting 3 from the second inequality gives us 5>3 and 3>-5. Combining these two inequalities gives us 5>-5, showing that a is greater than -c.

4. What is the significance of solving a>b, b>-c, and a>-c?

Solving this set of inequalities allows us to determine the relationship between three quantities. In this case, we can conclude that a is greater than -c, which can be useful in various mathematical and scientific applications.

5. Are there any other properties or rules that are important to solving these types of inequalities?

Yes, in addition to the transitive property, the addition and subtraction properties of inequality can also be used. These state that if A>B, then A+C>B+C and A-C>B-C. This can be helpful when working with more complex inequalities involving multiple variables.

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