Lesson 12
... The relation of provability (A1,...,An |– A) and the relation of logical entailment (A1,...,An |= A) are distinct relations. Similarly, the set of theorems |– A (of a calculus) is generally not identical to the set of logically valid formulas |= A. The former is a syntactic issue and defined within ...
... The relation of provability (A1,...,An |– A) and the relation of logical entailment (A1,...,An |= A) are distinct relations. Similarly, the set of theorems |– A (of a calculus) is generally not identical to the set of logically valid formulas |= A. The former is a syntactic issue and defined within ...
Chapter 1 - UTRGV Faculty Web
... Now, AD-BD = (A-B)D = k ND So, N divides AD-BD. Therefore, AD BD (mod N) ...
... Now, AD-BD = (A-B)D = k ND So, N divides AD-BD. Therefore, AD BD (mod N) ...
Class Notes (Jan.30)
... • A theorem is a mathematical statement that can be shown to be true. • An axiom or postulate is an assumption accepted without proof. • A proof is a sequence of statements forming an argument that shows that a theorem is true. The premises of the argument are axioms and previously proved theorems. ...
... • A theorem is a mathematical statement that can be shown to be true. • An axiom or postulate is an assumption accepted without proof. • A proof is a sequence of statements forming an argument that shows that a theorem is true. The premises of the argument are axioms and previously proved theorems. ...
Mathematical Logic and Foundations of
... undecidability of (N, +, ×, =) decidability of (R, +, ×, =) introduction to set theory axiomatic set theory the Axiom of Choice the Continuum Hypothesis ...
... undecidability of (N, +, ×, =) decidability of (R, +, ×, =) introduction to set theory axiomatic set theory the Axiom of Choice the Continuum Hypothesis ...
(A B) |– A
... Or, In the third case C1 = A, and we are to prove A A (see example 1). b) Induction step: we prove that on the assumption of A Cn being proved for n = 1, 2, ..., i-1 the formula A Cn can be proved also for n = i. For Ci there are four cases: 1. Ci is an assumption of Ai, 2. Ci is an axiom, 3. ...
... Or, In the third case C1 = A, and we are to prove A A (see example 1). b) Induction step: we prove that on the assumption of A Cn being proved for n = 1, 2, ..., i-1 the formula A Cn can be proved also for n = i. For Ci there are four cases: 1. Ci is an assumption of Ai, 2. Ci is an axiom, 3. ...
(A B) |– A
... The relation of provability (A1,...,An |– A) and the relation of logical entailment (A1,...,An |= A) are distinct relations. Similarly, the set of theorems |– A (of a calculus) is generally not identical to the set of logically valid formulas |= A. The former is a syntactic issue and defined within ...
... The relation of provability (A1,...,An |– A) and the relation of logical entailment (A1,...,An |= A) are distinct relations. Similarly, the set of theorems |– A (of a calculus) is generally not identical to the set of logically valid formulas |= A. The former is a syntactic issue and defined within ...
(1) Find all prime numbers smaller than 100. (2) Give a proof by
... (1) Find all prime numbers smaller than 100. (2) Give a proof by induction (instead of a proof by contradiction given in class) that any natural number > 1 has a unique (up to order) factorization as a product of primes. (3) Give a proof by induction that if a ≡ b( mod m) then an ≡ bn ( mod m) for a ...
... (1) Find all prime numbers smaller than 100. (2) Give a proof by induction (instead of a proof by contradiction given in class) that any natural number > 1 has a unique (up to order) factorization as a product of primes. (3) Give a proof by induction that if a ≡ b( mod m) then an ≡ bn ( mod m) for a ...
Week 13
... Exercise 5: As with previous proofs without words, explain the one given below. Exercise 6: The fact “proved” to the left can be demonstrated just as well by drawing a line segment (with length 1) and dividing it over and over again in much the same way. Show this version of the proof without words. ...
... Exercise 5: As with previous proofs without words, explain the one given below. Exercise 6: The fact “proved” to the left can be demonstrated just as well by drawing a line segment (with length 1) and dividing it over and over again in much the same way. Show this version of the proof without words. ...
1.4 Proving Conjectures: Deductive Reasoning
... 1.4 Proving Conjectures: Deductive Reasoning Today’s Goal: To prove mathematical statements using a logical argument. Deductive reasoning: __________________________________________________ __________________________________________________ Mathematical Proof Proof: _________________________________ ...
... 1.4 Proving Conjectures: Deductive Reasoning Today’s Goal: To prove mathematical statements using a logical argument. Deductive reasoning: __________________________________________________ __________________________________________________ Mathematical Proof Proof: _________________________________ ...
Math Review
... • Assume the theorem is false (so there are only finite prime) • Let P1, P2, ..., Pk be all the primes in increasing order. • Let N = P1P2 Pk + 1,N is > Pk , so it is not a prime • But it is also not divisible by any of the listed primes, contradicting the factorization of integers into prim ...
... • Assume the theorem is false (so there are only finite prime) • Let P1, P2, ..., Pk be all the primes in increasing order. • Let N = P1P2 Pk + 1,N is > Pk , so it is not a prime • But it is also not divisible by any of the listed primes, contradicting the factorization of integers into prim ...
Proofs - faculty.cs.tamu.edu
... A proof is a sequence of statements, each of which is either assumed, or follows follows from preceding statements by a rule of inference. We already learned many rules of inference (and essentially all of them are common sense rules). ...
... A proof is a sequence of statements, each of which is either assumed, or follows follows from preceding statements by a rule of inference. We already learned many rules of inference (and essentially all of them are common sense rules). ...
(A B) |– A
... if T, A |– B and |– A, then T |– B. It is not necessary to state theorems in the assumptions. if A |– B, then T, A |– B. (Monotonicity of proving) if T |– A and T, A |– B, then T |– B. if T |– A and A |– B, then T |– B. if T |– A; T |– B; A, B |– C then T |– C. if T |– A and T |– B, then T |– A B. ...
... if T, A |– B and |– A, then T |– B. It is not necessary to state theorems in the assumptions. if A |– B, then T, A |– B. (Monotonicity of proving) if T |– A and T, A |– B, then T |– B. if T |– A and A |– B, then T |– B. if T |– A; T |– B; A, B |– C then T |– C. if T |– A and T |– B, then T |– A B. ...
Chapter 1: The Foundations: Logic and Proofs
... – Case 2 . We show that if x2 is even then x must be even . We use an indirect proof: Assume x is not even and show x2 is not even. If x is not even then it must be odd. So, x = 2k + 1 for some k. Then x2 = (2k+1) 2 = 2(2k 2+2k)+1 which is odd and hence not even. This completes the proof of the seco ...
... – Case 2 . We show that if x2 is even then x must be even . We use an indirect proof: Assume x is not even and show x2 is not even. If x is not even then it must be odd. So, x = 2k + 1 for some k. Then x2 = (2k+1) 2 = 2(2k 2+2k)+1 which is odd and hence not even. This completes the proof of the seco ...
existence and uniqueness of binary representation
... Consider two cases. First, suppose that k is even. That is, there exists m < k such that k = 2m. In that case, since m satisfies the conditions of the induction hypothesis, k = 2(cr 2r + · · · + c0 ) = Σri=0 ci 2i+1 . It is easy to see that this formula gives us a binary representation of k with al ...
... Consider two cases. First, suppose that k is even. That is, there exists m < k such that k = 2m. In that case, since m satisfies the conditions of the induction hypothesis, k = 2(cr 2r + · · · + c0 ) = Σri=0 ci 2i+1 . It is easy to see that this formula gives us a binary representation of k with al ...
Quiz04-soln - Rose
... Induction step. IN(T) > 0, so T has a root (internal node), and two subtrees TL and TR. Notice that IN(T) + IN(TL) + IN(TR) + 1, and EN(T) = EN(TL) + EN(TR). Induction assumption. The property is true for any EBT with fewer nodes than T. In particular, for TL and TR, which are smaller because they d ...
... Induction step. IN(T) > 0, so T has a root (internal node), and two subtrees TL and TR. Notice that IN(T) + IN(TL) + IN(TR) + 1, and EN(T) = EN(TL) + EN(TR). Induction assumption. The property is true for any EBT with fewer nodes than T. In particular, for TL and TR, which are smaller because they d ...
Document
... two-valued logic – every sentence is either true or false some sentences are minimal – no proper part which is also a sentence others – can be taken apart into smaller parts we can build larger sentences from smaller ones by using connectives ...
... two-valued logic – every sentence is either true or false some sentences are minimal – no proper part which is also a sentence others – can be taken apart into smaller parts we can build larger sentences from smaller ones by using connectives ...
