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