Diccrete Math problem: finding a proposition given a specific truth table

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SUMMARY

The discussion focuses on deriving a logical proposition from a specific truth table using only the logical operators p, q, ¬ (NOT), and ∧ (AND). The truth table provided indicates that the resulting proposition is ¬p, achieved through systematic simplification of the expressions derived from the true evaluations of the table. The alternative answer presented from the book, ¬(p ∧ ¬q) ∧ ¬(p ∧ q), is also logically equivalent to the derived proposition, highlighting the flexibility in logical expression representation.

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  • Understanding of propositional logic and truth tables
  • Familiarity with logical operators: AND (∧), NOT (¬)
  • Knowledge of logical equivalence and simplification techniques
  • Experience with systematic approaches to solving logical problems
NEXT STEPS
  • Study logical equivalences in propositional logic
  • Learn about simplification techniques in Boolean algebra
  • Explore systematic methods for constructing logical propositions
  • Investigate the use of other logical operators, such as OR (∨), in truth table analysis
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Students of discrete mathematics, logic enthusiasts, and anyone interested in mastering propositional logic and truth table analysis.

nicnicman
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I've been working at this problem for a while and it seems that there should be an easier more systematic way of solving it. Here it is:

Find a proposition using only p, q, ¬ and the connective ∧ with the given truth table.

p q ?
T T F
T F F
F T T
F F T

I know of a systematic approach to creating propositions but includes the use of OR. Is there a systematic way of solving this only using ANDs?
 
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Just read off the ones where the truth table evaluates to "T", namely:
F T T
F F T
This results in:
(not p and q) (from the F T T line)
(not p and not q) (from the F F T line)

join them with "or":
(not p and q) or (not p and not q)

simplify if possible, using the distributive rule in this case:
(not p) and (q or not q)
q or not q = T
(not p) and T = not p

final result:
not p
 
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Thanks this is very helpful. It's actually the way I originally went about solving the problem but the book gives another answer:

¬(p ∧ ¬q) ∧ ¬(p ∧ q)

Probably both answers are acceptable, but I wonder how I could get the book's answer from your original answer, (¬p ∧ q) ∨ (¬p ∧ ¬q). There must be a way to convert it since they are logically equivalent.
 

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