¬(p∧q), prove (¬p∨¬q). This is done by a truth table. 1. Negation: Rule for ¬x is: If x has a T, then ¬x has an F and vice-versa. 2. Conjunction: Rule for x∧y is: If x and y both have T's, x∧y has T, otherwise x∧y has F. 3. Disjunction: Rule for x∨y is: If x and y both have F's, x∨y has F, otherwise x∨y has T. 4. Biconditional: Rule for x<->y is: If both x and y have the same truth value, x<->y has T, otherwise F. You need headings as listed below: p | q | ¬p | ¬q | p∧q | ¬(p∧q) | ¬p∨¬q | ¬(p∧q) <-> (¬p∨¬q) | --------------------------------------------------------------- T | T | F | F | T | F | F | T | T | F | | | | | | | F | T | | | | | | | F | F | | | | | | | The F under ¬p is because there is a T under p, rule 1 above. The F under ¬q is because there is a T under q, rule 1 above. The T under p∧q is because there is a T under both p and q, rule 2 above. The F under ¬(p∧q) is because there is a T under p∧q, rule 1 above. The F under ¬p∨¬q is because ¬p and ¬q both have F's rule 3 above. The T under ¬(p∧q) <-> (¬p∨¬q) is because both ¬(p∧q) and (¬p∨¬q) have the same truth value F, rule 4 Now we fil in the next row: p | q | ¬p | ¬q | p∧q | ¬(p∧q) | ¬p∨¬q | ¬(p∧q) <-> (¬p∨¬q) | --------------------------------------------------------------- T | T | F | F | T | F | F | T | T | F | F | T | F | T | T | T | F | T | | | | | | | F | F | | | | | | | The F under ¬p is because there is a T under p, rule 1 above. The T under ¬q is because there is a F under q, rule 1 above. The F under p∧q is because p and q don't both have T's, rule 2 above. The T under ¬(p∧q) is because there is a F under p∧q, rule 1 above. The T under ¬p∨¬q is because ¬p and ¬q don't both have F's rule 3 above. The T under ¬(p∧q) <-> (¬p∨¬q) is because both ¬(p∧q) and (¬p∨¬q) have the same truth value T, rule 4 Now we fil in the third row: p | q | ¬p | ¬q | p∧q | ¬(p∧q) | ¬p∨¬q | ¬(p∧q) <-> (¬p∨¬q) | --------------------------------------------------------------- T | T | F | F | T | F | F | T | T | F | F | T | F | T | T | T | F | T | T | F | F | T | T | T | F | F | | | | | | | The T under ¬p is because there is a F under p, rule 1 above. The F under ¬q is because there is a T under q, rule 1 above. The F under p∧q is because p and q don't both have T's, rule 2 above. The T under ¬(p∧q) is because there is a F under p∧q, rule 1 above. The T under ¬p∨¬q is because ¬p and ¬q don't both have F's rule 3 above. The T under ¬(p∧q) <-> (¬p∨¬q) is because both ¬(p∧q) and (¬p∨¬q) have the same truth value T, rule 4 Now we fill in the fourth row: p | q | ¬p | ¬q | p∧q | ¬(p∧q) | ¬p∨¬q | ¬(p∧q) <-> (¬p∨¬q) | --------------------------------------------------------------- T | T | F | F | T | F | F | T | T | F | F | T | F | T | T | T | F | T | T | F | F | T | T | T | F | F | T | T | F | T | T | T | The T under ¬p is because there is a F under p, rule 1 above. The T under ¬q is because there is a F under q, rule 1 above. The F under p∧q is because p and q don't both have T's, rule 2 above. The T under ¬(p∧q) is because there is a F under p∧q, rule 1 above. The T under ¬p∨¬q is because ¬p and ¬q don't both have F's rule 3 above. The T under ¬(p∧q) <-> (¬p∨¬q) is because both ¬(p∧q) and (¬p∨¬q) have the same truth value T, rule 4. Now since there are only T's in the last column, that proves that in every case, the biconditional ¬(p∧q) <-> (¬p∨¬q) is true and therefore ¬(p∧q) and (¬p∨¬q) are equivalent because the biconditional is a tautology. Edwin