
Modal Logic
... Every atomic proposition p is a member of . If is a member of , so is (¬). If and are members of , so is (). If is a member of , so is (❏). ...
... Every atomic proposition p is a member of . If is a member of , so is (¬). If and are members of , so is (). If is a member of , so is (❏). ...
Continuity and uniform continuity with epsilon and delta
... one of x, y is less than 1, then both are smaller than 2 (since δ ≤ 1) and then |f (x) − f (y)| < (since δ ≤ δ1 ). On the other hand, if x and y are both in [1, ∞), then < 2 < . This proves that f is uni|f (x) − f (y)| < |x−y| ...
... one of x, y is less than 1, then both are smaller than 2 (since δ ≤ 1) and then |f (x) − f (y)| < (since δ ≤ δ1 ). On the other hand, if x and y are both in [1, ∞), then < 2 < . This proves that f is uni|f (x) − f (y)| < |x−y| ...
Rewriting Equations
... Slides 4, 7, and 12 are the worked out examples, and slide 16 is the challenge problem. May want to consider printing those out for their notebook. We quiz Thurs., if you are behind you may want to print all of it out to save time. The challenge problem comes from p.189 #37 ...
... Slides 4, 7, and 12 are the worked out examples, and slide 16 is the challenge problem. May want to consider printing those out for their notebook. We quiz Thurs., if you are behind you may want to print all of it out to save time. The challenge problem comes from p.189 #37 ...
Bound and Free Variables Theorems and Proofs
... using a random process, that puts each edge in with probability 1/2. ...
... using a random process, that puts each edge in with probability 1/2. ...
Slides
... Let be an unsatisfiable formula Theorem 2: A proof P is complete with respect to if for every full assignment ...
... Let be an unsatisfiable formula Theorem 2: A proof P is complete with respect to if for every full assignment ...
Handout 14
... where the Ai can be either atomic formulas or their negations. It is thus a conjuction of clauses, i.e. disjunctions of literals (variables or their negations). CNF is used in machine proving of theorems. Resolution in Prolog is also based on a special form of CNF. Definition 5.10 (Disjunctive norma ...
... where the Ai can be either atomic formulas or their negations. It is thus a conjuction of clauses, i.e. disjunctions of literals (variables or their negations). CNF is used in machine proving of theorems. Resolution in Prolog is also based on a special form of CNF. Definition 5.10 (Disjunctive norma ...