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Hi!
I'm having some trouble understanding how I'm supposed to use the reduction and deduction methods.
The question is:
(p → q) ∧ (¬p → r) ∧ ((¬p ∧ r) → s) ∧ ¬q ⇒ s
Prove that this is correct, with the deduction AND reduction method.
I'm fairly new to this kind mathematics, so if somebody could explain and guide me through the answer I would be really grateful!
PS: Excuse the bad grammar (if any), english is my fourth language.
Thanks in advance.
Greetings!
So the given implication is correct.
Bassaricyon neblina
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Hi Olanguito and thank you for the reply!
I've done the following:
(p → q) ∧ (¬p → r) ∧ ((¬p ∧ r) → s) ∧ ¬q ⇒ s
1. ¬q (condition)
2. p → q (-II-)
3. ¬p (Modus Tollens)
4. ¬p → r (condition)
5. pvr (equivalence) <=> ¬p ∧ r
6. r (disjunctive syllogism)
7. ¬p ∧ r (conjunction)
8. (¬p ∧ r) → s (condition)
9. s (Modus ponens)
1. q can not be true because ¬q is true.
2. s is false, which means ¬p ∧ r is also false.
3. p is false because q is false.
4. r is true because ¬p is true.
5. ((¬p ∧ r) → is false, which gives us an divergence.
(p → q) ∧ (¬p → r) ∧ ((¬p ∧ r) → s) ∧ ¬q ⇒ s
5. 2. 2. 6. 4. 2. 2.3.1.2
0. 1. 1. 1. 0. 1. 1.0.0.0
To all of this I got the following response:
- Adoption missing.
- The figures in the expression can not keep up with your numbers in the text below.
- Not clear why point 5 and what is the contradiction.
- Conclusion (what is contradicted really?)
Could somebody tell me how I can "fix" these problems. I'm kinda new to the reduction- and deduction methods so I was basically using the notes from my lectures and Google as a reference. Probably why I managed to screw things up. But it looks like I'm halfway to finishing it? Where did I go wrong in my thinking?
Thanks in advance!
Greetings!
Last edited by MogiYagi (2014-10-07 03:42:23)
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