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Mathematical Logic - Propositional Logic
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p \oplus q
True if p and q have different truth values, false if they are the same.
p \lor q
True if at least one of p or q is true, otherwise false.
p \rightarrow q
If p is true, then q is also true. If p is false, then nothing is said about q.
p \leftrightarrow q
True if p and q have the same truth value, false otherwise.
(p \lor q) \rightarrow r
True if whenever at least one of p or q is true, r is also true.
(p \land q) \rightarrow r
True if whenever both p and q are true, r is also true.
p \land q
True if both p and q are true, otherwise false.
p \rightarrow (q \land r)
True if whenever p is true, both q and r are also true.
\neg p
True if p is false, and false if p is true.
p \rightarrow (q \rightarrow p)
True, this implication always holds due to the reflexivity of implication.
((p \rightarrow q) \land (q \rightarrow r)) \rightarrow (p \rightarrow r)
True, as this is the transitive property of implications.
\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 \lor q)
True if both p and q are false.
(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) \land (p \rightarrow r)
True if p is true and both q and r are also true, or if p is false.
(p \lor q) \land \neg(p \and q)
True if exactly one of p or q is true (exclusive or).
(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.
\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 \lor (q \land r)) \leftrightarrow ((p \lor q) \and (p \lor r))
True, this represents the distributive law of disjunction over conjunction.
((p \lor q) \land \neg p) \rightarrow q
True, as this represents the disjunctive syllogism rule of inference.
(p \rightarrow r) \land (q \rightarrow r)
True if either both p and q are false, or if r 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 \land q)
True if either p is false, q is false, or both are false.
((p \rightarrow q) \land p) \rightarrow q
True, direct application of the modus ponens inference rule.
\neg p \rightarrow (p \rightarrow q)
True because if p is false then the implication 'p implies q' is always true.
(p \land q) \leftrightarrow (q \land p)
True, as the conjunction operation is commutative.
\neg p \rightarrow (p \lor q)
True, based on the principle that the disjunction is true if at least one disjunct is true.
p \rightarrow (q \lor r)
True if whenever p is true, at least one of q or r is true.
\neg (p \leftrightarrow q) \leftrightarrow (p \oplus q)
True, this shows that the negation of biconditional is equivalent to exclusive or.
(p \lor q) \leftrightarrow (q \lor p)
True, as the disjunction operation is commutative.
(p \rightarrow q) \leftrightarrow (\neg q \rightarrow \neg p)
True, this is known as contraposition.
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