true, Q must be true. P Q P IMPLIES Q true true true true false false false true true false false true This truth table defines implies: there is no implied causality or connection with what implies implies in English!
impact signer’s behavior • Varying impact (?): – Quill on parchment – Pen on paper – Scribbling on tablet – Typing – Clicking a web form Purposeful Signatures • Verifiable • Non-‐repudiable • Bound to content
electronic signatures: One-‐way hard problem Easy to raise to powers, hard to find discrete logs Signature combines message and private key Can be verified by obtaining public key from trusted source and checking signature is valid
least one of x or y must be even. P = “the product of x and y is even” Q = “at least one of x or y must be even” What definitions do we need? Goal: prove P implies Q
least one of x or y must be even. P = “the product of x and y is even” Q = “at least one of x or y must be even” Goal: prove P implies Q Definition: even. An integer, z, is even if there exists an integer k such that z = 2k.
least one of x or y must be even. P = “the product of x and y is even” Q = “at least one of x or y must be even” Goal: prove P implies Q Definition: even. An integer, z, is even if there exists an integer k such that z = 2k.
Q = “at least one of x or y must be even” To prove, P implies Q, we use contrapositive inference rule: NOT(Q) IMPLIES NOT(P) P IMPLIES Q Observe: this is starting backwards! We are starting the proof from the conclusion we want.
Q = “at least one of x or y must be even” To prove, P implies Q, we prove the contrapositive: NOT(Q) implies NOT(P) To prove an implication, assume left side, show right: Assume NOT(at least one of x and y must be even)
Q = “at least one of x or y must be even” To prove, P implies Q, we prove the contrapositive: NOT(Q) implies NOT(P) To prove an implication, assume left side, show right: Assume NOT(at least one of x and y must be even) By the meaning of NOT: both x and y are not even
if there exists an integer k such that z = 2k. Definition: odd. An integer, z, is odd if and only if there exists an integer k such that z = 2k + 1. Odd-‐Even Lemma: If an integer is not even, it is odd.
Q = “at least one of x or y must be even” To prove, P implies Q, we prove the contrapositive: NOT(Q) implies NOT(P) To prove an implication, assume left side, show right: Assume NOT(at least one of x and y must be even) By the meaning of NOT: both x and y are not even By the Odd-‐Even Lemma: both x and y are odd
Q = “at least one of x or y must be even” To prove, P implies Q, we prove the contrapositive: NOT(Q) implies NOT(P) To prove an implication, assume left side, show right: Assume NOT(at least one of x and y must be even) By the meaning of NOT: both x and y are not even By the Odd-‐Even Lemma: both x and y are odd By the definition of odd: there exists integers k, m, such that x = 2k + 1 and y = 2m + 1
Q = “at least one of x or y must be even” We prove the contrapositive: Assume NOT(at least one of x and y must be even) By the meaning of NOT: both x and y are not even. By the Odd-‐Even Lemma: both x and y are odd. By the definition of odd: there exists integers k, m, such that x = 2k + 1 and y = 2m + 1 By algebra: xy = (2k + 1)(2m + 1) = 4mk + 2m + 2k + 1 = 2(2mk + m + k) + 1
Q = “at least one of x or y must be even” We prove the contrapositive: Assume NOT(at least one of x and y must be even) … Since the integers closed under multiplication and addition: there exists some integer r where r = 2mk + m + k By definition of odd: So xy = 2r + 1 which means the product of x and y is odd. By the Odd-‐Even Lemma: the product of x and y is not even.
Q = “at least one of x or y must be even” We prove the contrapositive: Assume NOT(at least one of x and y must be even) … By integers closed under multiplication and addition: there exists some integer r where r = 2mk + m + k By definition of odd: So xy = 2r + 1 which means the product of x and y is odd. By the Odd-‐Even Lemma: the product of x and y is not even. Odd-‐Even Lemma: If an integer is not even, it is odd.
Q = “at least one of x or y must be even” We prove the contrapositive: Assume NOT(at least one of x and y must be even) … By the Even-‐Odd* Lemma: the product of x and y is not even. So, NOT(product of x and y is not even) Thus, we have proven the implication: NOT(at least one of x and y must be even) implies NOT(at least one of x and y must be even)
Q = “at least one of x or y must be even” We prove the contrapositive: Assume NOT(at least one of x and y must be even) … So, NOT(product of x and y is not even) Thus, we have proven the implication: NOT(at least one of x and y must be even) implies NOT(at least one of x and y must be even) By the contrapositive inference rule, this proves: at least one of x and y must be even implies at least one of x and y must be even.
and accepted Each step uses a sound inference rule correctly: – Shows antecedents are satisfied – Concludes the conclusion Results in concluding goal proposition: P Do the proofs we do in cs2102 actually do this?
– Should be obvious what you are proving and how • Convincing to a skeptical reader • State assumptions clearly: careful about not assuming non-‐obvious things • Focus on important steps, not gory details
Before Friday (6:29pm): course pledge, registration survey (read “Habits of Highly Mathematical People”) • Read MCS Ch 2, 3; PS1 is due next Friday • TA’s office hours will start next week (and be posted soon)