Question 1009929: Natural Deduction - all 18 rules can be used
1. M -> (R ^ E)
2. (E v H) -> G / M -> G
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Natural Deduction
| Number | Statement | Lines Used | Reason |
|---|
| 1 | M -> (R ^ E) | | | | 2 | (E v H) -> G | | | | :. | M -> G | | | | 3 | ~(E v H) v G | 2 | MI | | 4 | (~E ^ ~H) v G | 3 | DM | | 5 | G v (~E ^ ~H) | 4 | Comm | | 6 | (G v ~E) ^ (G v ~H) | 5 | Dist | | 7 | G v ~E | 6 | Simp | | 8 | ~E v G | 7 | Comm | | 9 | E -> G | 8 | MI | | 10 | ~M v (R ^ E) | 1 | MI | | 11 | (~M v R) ^ (~M v E) | 10 | Dist | | 12 | (~M v E) ^ (~M v R) | 11 | Comm | | 13 | ~M v E | 21 | Simp | | 14 | M -> E | 13 | MI | | 15 | M -> G | 14,9 | HS |
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Conditional Proof (alternative method)
Assume M is true. Show this leads to G being true.
| Number | Statement | Lines Used | Reason |
|---|
| 1 | | M -> (R ^ E) | | | | 2 | | (E v H) -> G | | | | :. | | M -> G | | | | 3 | M | | ACP | | 4 | R ^ E | 1,3 | MP | | 5 | E ^ R | 4 | Comm | | 6 | E | 5 | Simp | | 7 | E v H | 6 | Add | | 8 | G | 2,7 | MP | | 9 | | M -> G | 3-8 | CP |
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Abbreviations/Acronyms Used (applies to either method)
ACP = Assumption for Conditional Proof
Add = Addition
Comm = Commutation
CP = Conditional Proof
Dist = Distribution
DM = De Morgan's Law
HS = Hypothetical Syllogism
MI = Material Implication
MP = Modus Ponens
Simp = Simplification
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