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

[drema9]

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.

 

[mcampbell]

I'm confused by the first line. Is it A\rightarrowB and B\rightarrowC ?

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.

 

[drema9]

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.

 

[mcampbell]

I see... In your original post you have b>c. Your deduction is correct.

 

[drema9]

cool

cool,
any idea what a next step would be? this one is driving me nuts

 

[Link-]

I think you only need to find -A and you can make an Add to get -A v -C.

 

[Link-]

the question is.
1. (a>b). (b>c)

I don't understand the (.) operator, What does mean?

 

[Link-]

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\rightarrowQ:: -P v Q

 

[drema9]

ok, i am still stuck i can not get -a then i can wedge in the last part.

 

[Link-]

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

 

[MathematicalPhysicist]

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!