Having problems with solving this question. The directions are: Use the 18 rules of inference, supply a proof.
1.(RvX)>(A>B)
2.~Q>~C
3.~C>Z
4.R.Y
5.QvA
The conclusion I'm supposed to arrive at is /ZvB
To solve this question using the 18 rules of inference, you need to apply these rules step by step to reach the desired conclusion: /ZvB. Here's a step-by-step explanation of how to solve this proof:
1. (RvX) > (A > B) - Given
2. ~Q > ~C - Given
3. ~C > Z - Given
4. R • Y - The dot (∙) represents the conjunction, also known as the "and" operator. This means R and Y are both true.
5. Q v A - Given
To arrive at the conclusion /ZvB, we need to break this into two separate subconclusions: Z and B. We will use two separate subproofs to derive each of them. Here's how it's done:
Proving Z:
6. Assume ~Z (subproof starts)
7. ~C - Line 3, Modus Ponens
8. ~Z v B - Disjunction Introduction (DI) on Line 7
9. Assume Z (subproof starts)
10. Z - Line 9, Assumption
11. Z v B - Disjunction Introduction (DI) on Line 10
12. Z v B - Disjunction Elimination (DE) on Lines 6-8 and 9-11
13. /ZvB - Negation Introduction (NI) on Line 6 and subproofs from 9-12
Proving B:
14. Assume ~B (subproof starts)
15. ~B v (A > B) - Disjunction Introduction (DI) on Line 14
16. ~R v ~X - De Morgan's Law on Line 1 and 15
17. ~(R • Y) - De Morgan's Law on Line 16
18. R • Y - Line 4
19. Contradiction: ~(R • Y) and (R • Y) are both true (Lines 17 and 18)
20. B - Negation Elimination (NE) on Line 14 and the contradiction on Line 19
21. Z v B - Disjunction Introduction (DI) on Line 20
22. /ZvB - Negation Introduction (NI) on Line 14 and subproofs from 15-21
As a result, by deriving both Z and B through separate subproofs, we can conclude /ZvB.