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Mathematical Logic - Propositional Logic
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p \land q
True if both p and q are true, otherwise false.
(p \lor q) \rightarrow r
True if whenever at least one of p or q is true, r is also true.
p \rightarrow (q \land r)
True if whenever p is true, both q and r are also true.
p \lor q
True if at least one of p or q is true, otherwise false.
\neg p
True if p is false, and false if p is true.
(p \land q) \rightarrow r
True if whenever both p and q are true, r is also true.
p \rightarrow q
If p is true, then q is also true. If p is false, then nothing is said about q.
p \oplus q
True if p and q have different truth values, false if they are the same.
p \leftrightarrow q
True if p and q have the same truth value, false otherwise.
p \rightarrow (q \lor r)
True if whenever p is true, at least one of q or r is true.
(p \land (p \rightarrow q)) \rightarrow q
This is a form of modus ponens: true if q is true whenever p is true and 'p implies q' is true.
(p \rightarrow q) \leftrightarrow (\neg q \rightarrow \neg p)
True, this is known as contraposition.
(p \lor q) \leftrightarrow (q \lor p)
True, as the disjunction operation is commutative.
\neg(\neg p)
True if p is true, and false if p is false. It is simply the value of p.
(\neg q \land (p \rightarrow q)) \rightarrow \neg p
This is a form of modus tollens: true if p is false whenever q is false and 'p implies q' is true.
(p \land (p \rightarrow q)) \leftrightarrow q
Not necessarily true; equivalent to modus ponens but represented as a biconditional, which requires equivalence.
((p \rightarrow q) \land (q \rightarrow r)) \rightarrow (p \rightarrow r)
True, as this is the transitive property of implications.
(p \rightarrow r) \land (q \rightarrow r)
True if either both p and q are false, or if r is true.
(p \lor q) \land \neg(p \and q)
True if exactly one of p or q is true (exclusive or).
\neg(p \lor q)
True if both p and q are false.
\neg (p \rightarrow q) \leftrightarrow (p \land \neg q)
True, this is based on the definition of implication and represents the only case where 'p implies q' is false.
\neg (p \leftrightarrow q) \leftrightarrow (p \oplus q)
True, this shows that the negation of biconditional is equivalent to exclusive or.
((p \lor q) \land \neg p) \rightarrow q
True, as this represents the disjunctive syllogism rule of inference.
(p \rightarrow q) \land (p \rightarrow \neg q)
True only if p is false, because q and \neg q cannot both be true.
(p \lor \neg p)
True, this is known as the law of excluded middle.
(p \land q) \leftrightarrow (q \land p)
True, as the conjunction operation is commutative.
\neg(p \land q)
True if either p is false, q is false, or both are false.
p \rightarrow (q \rightarrow p)
True, this implication always holds due to the reflexivity of implication.
(p \rightarrow q) \land (p \rightarrow r)
True if p is true and both q and r are also true, or if p is false.
((p \rightarrow q) \land p) \rightarrow q
True, direct application of the modus ponens inference rule.
\neg p \rightarrow (p \lor q)
True, based on the principle that the disjunction is true if at least one disjunct is true.
(p \land (q \lor r)) \leftrightarrow ((p \land q) \lor (p \land r))
True, this represents the distributive law of conjunction over disjunction.
\neg p \rightarrow (p \rightarrow q)
True because if p is false then the implication 'p implies q' is always true.
(p \lor (q \land r)) \leftrightarrow ((p \lor q) \and (p \lor r))
True, this represents the distributive law of disjunction over conjunction.
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