If statement implies second statement, first statement implies both itself and second statement: If (p -> q), then p -> (p & q) {absorption rule}.
If statement is true, it implies the statement that either the statement is true and/or second statement is true: If p, then (p | q) {addition rule}.
a & (b & c) = (a & b) & c. a | (b | c) = (a | b) | c {association rule}|.
Statements can connect by IF AND ONLY IF ... THEN ..., as in IF AND ONLY IF a THEN b or IFF a THEN b {biconditional} {iff}. If and only if means theorem and converse. Biconditional is true if and only if both statements are true or both statements are false.
a & b = b & a. a | b = b | a {commutation rule}|.
First statement implies second statement AND third statement implies fourth statement. First statement OR third statement. THEN second statement OR fourth statement {complex constructive dilemma}: (p -> q) & (r -> s). p | r. Therefore, q | s.
First statement implies second statement AND third statement implies fourth statement. NOT second statement OR NOT fourth statement. THEN NOT first statement OR NOT third statement {complex destructive dilemma}: (p -> q) & (r -> s). ~q | ~s. Therefore, ~p | ~r.
No statement can be both true and false {principle of contradiction} {contradiction principle}|.
In algebra of sets, 1 - (x + y) = (1 - x) (1 - y) and 1 - xy = (1 - x) + (1 - y) {De Morgan's laws, logic} {De Morgan laws, logic}. In propositional logic, not (x and y) equals not x or not y, and not (x or y) equals not x and not y: ~(x + y) = ~x - ~y, and ~(x - y) = ~x + ~y.
If A1, A2, ..., and An are true, then B is true. A1 is true. A2 is true. ... An-1 is true. If An is true, then B is true {deduction theorem}.
Statements can connect by OR (a OR b), where OR is inclusive {inclusive OR, disjunction} {disjunction, logic}| {alternation}. The only false disjunction is if both statements are false. OR can also mean a or b but not both a and b {exclusive OR}.
If first or second statement is true and second statement is not true, first statement is true: (p | q) & ~p, so q {disjunction rule}|.
a | (b & c) = (a | b) & (a | c) and a & (b | c) = (a & b) | (a & c) {distribution rule}|.
Negative of a negative statement is statement: ~(~p) = p {double negation}.
Statements are either true or false {excluded middle, principle}| {principle of the excluded middle}. Disjunction of statement and negative statement is true.
The statement that first statement and second statement imply third statement is materially equivalent to the statement that first implies second, which implies third: (p & q) -> r = p -> q -> r {exportation rule}. Exportation is true in propositional calculus. Exportation is not true for strict implication or entailment.
Statements imply that they are true statements {principle of identity} {identity principle}|. If statement, statement is true.
True statements imply true statements {implication, logic rule}|. False statements imply any statement.
Conclusion is equivalent to negative of conjunction of premise and negative of conclusion {material implication}|: b = -(a & -b). Material implication has sideways horseshoe or arrow -> symbol.
If first statement is true and statement that first statement implies second statement is true, second statement is true {modus ponens, rule}| {detachment rule} {rule of detachment} {affirming the antecedent}. If A is true, and A then B is true, then B is true. p & (p -> q) -> q. Modern formal logic requires only modus-ponens rule.
If A then B is true, and not-B is true, then not-A is true {modus tollens}| {denying the consequent}. If first statement implies second statement is true, and second statement is not true, then first statement is not true: (p -> q) & -q, so -p {principle of modus tollens}.
Most A is B {most statement} {statement using most}. Most A is C. Therefore, Some B are C.
First statement implies second statement AND third statement implies second statement. First statement OR third statement. THEN second statement {simple constructive dilemma}: (p -> q) & (r -> q). p | r. Therefore, q.
First statement implies second statement AND first statement implies third statement. NOT second statement OR NOT third statement. THEN NOT first statement {simple destructive dilemma}: (p -> q) & (p -> s). ~q | ~s. Therefore, ~p.
If the statement "first statement AND second statement" is true, first statement is true {simplification rule}: (p & q), so p.
P implies Q if and only if it is impossible that P is true and Q false {strict implication}| {logical implication}: (p -> q) = ~(p & ~q).
If first statement implies second statement, then second-statement negative implies first-statement negative {transposition rule}|: if (p -> q), then (-q -> -p).
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Date Modified: 2022.0225