Logic and Proofs

John

The following are some notes taken during a lecture.

Rules of Inference

  • We have the two premises:
    • “All men are mortal.”
    • “Socrates is a man.”
  • And the conclusion: 
    • “Socrates is mortal.”
  • How do we get the conclusion from the premises?

 

Rules of Inference: The Argument

  • We can express the premises (above the line) and the conclusion (below the line) in predicate logic as an argument:
  • We will see shortly that this is a valid argument.

 

Rules of Inference: Valid Arguments

  • We will show how to construct valid arguments in two stages; first for propositional logic and then for predicate logic. The rules of inference are the essential building block in the construction of valid arguments. 
    • Propositional Logic
      • Inference Rules
    • Predicate Logic
      • Inference rules for propositional logic plus additional inference rules to handle variables and quantifiers.

 

Arguments in Propositional Logic

  • An argument in propositional logic is a sequence of propositions. All but the final proposition are called premises. The last statement is the conclusion. 
  • The argument is valid if the premises imply the conclusion.  An argument form   is  an argument that is valid no matter what propositions are substituted into its propositional variables.    
  • If the premises are  p1 ,p2, …,pn  and the conclusion is q  then (p1  ∧ p2 ∧ … ∧ pn ) → q  is a tautology. 
  • Inference rules are simple argument forms that will be used to construct more complex argument forms.

 

Rules of Inference in Propositional Logic: Modus Ponens

  • Example:
    • Let p be “It is snowing.”
    • Let q be “I will study discrete math.”
    • “If it is snowing,  then I will study discrete math.”
    • “It is snowing.”
    • “Therefore , I will  study discrete math.”

 

Rules of Inference in Propositional Logic: Modus Tollens

  • Example:
    • Let p be “it is snowing.”
    • Let q be “I will study discrete math.”
    • “If it is snowing,  then I will study discrete math.”
    • “I will not study discrete math.”
    • “Therefore , it is not snowing.”

 

Rules of Inference in Propositional Logic: Hypothetical Syllogism

  • Example:
    • Let p be “it snows.”
    • Let q be “I will study discrete math.”
    • Let r be “I will get an A.”
    • “If it snows,  then I will study discrete math.”
    • “If I study discrete math, I will get an A.”
    • “Therefore , If it snows, I will get an A.”

 

Rules of Inference in Propositional Logic: Disjunctive Syllogism

  • Example:
    • Let p be “I will study discrete math.”
    • Let q be “I will study English literature.”
    • “I will study discrete math or I will study English literature.”
    • “I will not study discrete math.”
    • “Therefore , I will study English literature.”

 

Rules of Inference in Propositional Logic: Addition

  • Example:
    • Let p be “I will study discrete math.”
    • Let q be “I will visit Las Vegas.”
    • “I will study discrete math.”
    • “Therefore, I will  study discrete math or I will visit
    • Las Vegas.”

 

Rules of Inference in Propositional Logic: Simplification

  • Example:
    • Let p be “I will study discrete math.”
    • Let q be “I will study English literature.”
    • “I will study discrete math and English literature”
    • “Therefore, I will study discrete math.”

 

Rules of Inference in Propositional Logic: Conjunction

  • Example:
    • Let p be “I will study discrete math.”
    • Let q be “I will study English literature.”
    • “I will study discrete math.”
    • “I will study  English literature.”
    • “Therefore, I will study discrete math and I will study English literature.”

 

Rules of Inference in Propositional Logic: Resolution

  • Example:
    • Let p be “I will study discrete math.”
    • Let r be “I will study English literature.”
    • Let q be “I will study databases.”
    • “I will not study discrete math or I will study English literature.”
    • “I will study  discrete math or I will study databases.”
    • “Therefore, I will study databases or I will study English literature.”

 

 

Rules of Inference and Valid Arguments

  • A  valid argument is a sequence of statements. Each statement is either a premise or follows from previous statements by  rules of inference. The last statement is called conclusion.
  • A valid argument takes the following form:
    • S1
    • S2
    • .
    • .
    • .
    • Sn
    • therefore C

 

Valid Arguments

 

Example 2:

  • With these hypotheses:
    • “It is not sunny this afternoon and it is colder than yesterday.”
    • “We will go swimming only if it is sunny.”
    • “If we do not go swimming, then we will take a canoe trip.”
    • “If we take a canoe trip, then we will be home by sunset.”
  • Using the inference rules, construct a valid argument for the conclusion:
    • “We will be home by sunset.”
  • Solution:
    • Choose propositional variables:
      • p : “It is sunny this afternoon.”      r  : “We will go swimming.”  t : “We will be home by sunset.”
      • q  : “It is colder than yesterday.”     s  : “We will take a canoe trip.” 
    • Translation into propositional logic:

 

Handling Quantified Statements

  • Valid arguments for quantified statements are a sequence of statements. Each statement is either a premise or follows from previous statements by  rules of inference which include:
    • Rules of Inference for Propositional Logic
    • Rules of Inference for Quantified Statements
  • The rules of inference for quantified statements are introduced in the next several slides.

 

Universal Instantiation

  • Example:
    • Our domain consists of all dogs and Fido is a dog.
    • “All dogs are cuddly.”
    • “Therefore,  Fido is cuddly.”

 

Universal Generalization

  • Used often implicitly in Mathematical Proofs. 

 

Existential Instantiation

  • Example:
    • “There is someone who got an A in the course.”
    • “Let’s call her a and say that a got an A”

 

Existential Generalization

  • Example:
    • “Michelle got an A in the class.”
    • “Therefore,  someone got an A in the class.”

 

Using Rules of Inference

  • Example 1: Using the rules of inference, construct a valid argument to show that “John Smith has two legs” is a consequence of the premises: “Every man has two legs.” “John Smith is a man.”
  • Solution: Let M(x) denote  “x is a man” and L(x) “ x has two legs” and let John Smith be a member of the domain.
  • Valid Argument:

 

  • Example 2: Use the rules of inference to construct a valid argument showing that the conclusion “Someone who passed the first exam has not read the book.” follows from the premises
    • “A student in this class has not read the book.”
    • “Everyone in this class passed the first exam.”
  • Solution: Let C(x) denote  “x is in this class,” B(x) denote  “ x has  read the book,” and P(x) denote   “x passed the first exam.”
    •  First we translate the premises and conclusion into symbolic form.

 

Returning to Socrates example:

 

Rules of Inference

  • Universal Modus Ponens combines universal instantiation and modus ponens into one rule. 
  •  This rule could be used in the Socrates example.

 

 

Proof Techniques

Proofs of Mathematical Statements

  • A proof is a valid argument that establishes the truth of a statement.
  • In math, CS,  and other disciplines, informal proofs  which are generally shorter, are generally used.
    • More than one rule of inference are often used in a step. 
    • Steps may be skipped.
    • The rules of inference used are not explicitly stated. 
    • Easier for to understand and to explain to people. 
    • But it is also easier to introduce errors. 
  • Proofs have many practical applications:
    • verification that computer programs are correct 
    • establishing that operating systems are secure 
    • enabling programs to make inferences in artificial intelligence 
    • showing that system specifications are consistent

 

Some Definitions

  • A theorem is a statement that can be shown to be true using:
    • definitions
    • other theorems
    • axioms (statements which are given as true) 
    • rules of inference
  • A lemma is a ‘helping theorem’ or a result which is needed to prove a theorem.
  • A corollary is a result which follows directly from a theorem.
  • Less important theorems are sometimes called propositions. 
  • A conjecture is a statement that is being proposed to be true. Once a proof of a conjecture is found, it becomes a theorem. It may turn out to be false. 

 

Forms of Theorems

  • Many theorems assert that a property holds for all elements in a domain, such as the integers, the real numbers, or some of the discrete structures that we will study in this class. 
  • Often the universal quantifier (needed for a precise statement of a theorem) is omitted by standard mathematical convention.
    • For example, the statement: “If x > y, where x and y are positive real numbers, then x2 > y2 ” really means “For all positive real numbers x and y, if x > y, then x2 > y2 .”

 

Proving Theorems

  • Many theorems have the form:  
  • To prove them, we show that where c is an arbitrary element of the domain
  • By universal generalization the truth of the original formula follows.
  • So, we must prove something of the form: 

 

Proving Conditional Statements

  • Trivial Proof: If we know q is true, then p → q   is true as well.   
    • “If it is raining  then 1=1.”
  • Vacuous Proof: If we know p is false then  p → q   is true as well.
    • “If I am both rich and poor then 2 + 2 = 5.”

 

Proving Theorems: Even and Odd Integers

Definition:  The integer n is even if there exists an integer k such that n = 2k, and n is odd if there exists an integer k, such that n = 2k + 1. Note that every integer is either even or odd and no integer is both even and odd.

 

Proving Conditional Statements

  • Direct Proof: Assume that  p  is true. Use rules of inference, axioms, and logical equivalences to show that   q  must also be true.
  • Example: Give a direct proof of the theorem “If n is an odd integer, then n2  is odd.”
  • Solution: Assume that n is odd. Then n = 2k + 1 for an integer k. Squaring both sides of the equation, we get:
    • n2   = (2k + 1)2   = 4k2 + 4k +1 = 2(2k2 + 2k) + 1= 2r + 1,
    • where r = 2k2 + 2k , an integer.                                 
  • We have proved that if n is an odd integer, then n2  is an odd integer.    

 

  • Definition: The real number r is rational if there exist integers p and q where  q≠0  such that r = p/q
  • Example: Prove that the sum of two rational numbers is rational.
  • Solution: Assume r and s are two rational numbers. Then there must be integers p, q and also t, u  such that
  • Thus the sum is rational.

 

Proof by Contraposition

  • Proof by Contraposition: Assume ¬q  and show ¬p  is true also. This is sometimes called an indirect proof method. If we give a direct proof of ¬q → ¬p then we have a proof of p → q
    • Why does this work?
  • Example: Prove that if n is an integer and 3n + 2 is odd, then n is odd.
  • Solution: Assume n is even. So, n = 2k for some integer k. Thus 3n + 2 = 3(2k) + 2 =6k +2 = 2(3k + 1) = 2j  for j = 3k +1
    • Therefore 3n + 2 is even. Since we have shown ¬q → ¬p ,  p → q  must hold as well. If n is an integer and 3n + 2 is odd (not even) , then n is odd (not even).

 

  • Example: Prove that for an integer n, if n2  is odd, then n is odd.
  • Solution:  Use proof by contraposition. Assume n is even (i.e., not odd). 
    • Therefore, there exists an integer k such that n = 2k. Hence, n2   =  4k2 = 2 (2k2) and n2   is even(i.e., not odd).
    • We have shown that if n is an even integer, then n2  is even. Therefore by contraposition, for an integer n, if n2  is odd, then n is odd. 

 

Proving Conditionals

  • Proof by Contradiction: (AKA reductio ad absurdum). To prove  p, assume  ¬p  and derive a contradiction such as    p ∧ ¬p. (an indirect form of proof). Since we have shown that ¬p →F is true , it follows that the contrapositive  T→p also holds.
  • Example:  Prove that if you pick 22 days from the calendar, at least 4 must fall on the same day of the week.
  • Solution: Assume that no more than 3  of the 22 days fall on the same day of the week. Because there are 7 days of the week, we could only have picked 21 days. This contradicts the assumption that we have picked 22 days.

 

Proof by Contradiction

  • Example: Use a proof by contradiction to give a proof that  √2 is irrational.
  • Solution: Suppose √2 is rational. Then there exists integers a and b with √2  = a/b, where b≠ 0 and a and b have no common factors. Then
  • Therefore a2  must be even. If a2  is even then a must be even (an exercise). Since a is even, a = 2c  for some integer c. Thus,
  • Therefore b2  is even.  Again then b must be even as well.
  • But then 2 must divide both a and b. This contradicts our assumption that a and b have no common factors. We have proved by contradiction  that our initial assumption must be false  and  therefore  √2 is  irrational

 

Proof by Contradiction

  • Example: Prove that there is no largest prime number.
  • Solution: Assume that there is a largest prime number. Call it pn. Hence, we can list all the primes 2,3,.., pn. Form
  • None of the prime numbers on the list divides r. Therefore, either r is prime or there is a smaller prime that divides r. This contradicts the assumption that there is a largest prime. Therefore, there is no largest prime.

 

Proving Biconditionals

  • To prove a theorem that is a biconditional statement, that is, a statement of the form p ↔ q, we show that     p → q and q →p are both true.
  • Example: Prove the theorem: “If n is an integer, then n is odd if and only if n2  is odd.”
  • Solution:  We have already shown (previous slides) that both p →q and q →p. Therefore we can conclude p ↔ q.
    • Sometimes iff   is used as an abbreviation for “if an only if,” as in “If n is an integer, then n is odd iif n2  is odd.”

 

 

Proof by Cases

  • Example: Let  a @ b = max{a, b} = a  if a ≥ b,  otherwise a @ b = max{a, b} = b.
    • Show that for all  real numbers a, b, c
      • (a @b) @ c = a @ (b @ c)
      • (This means the operation @ is associative.)
  • Proof: Let a, b, and c be arbitrary real numbers. Then one of the following 6 cases must hold.
    • a ≥ b ≥ c
    • a ≥ c ≥ b
    • b ≥ a ≥c
    • b ≥ c ≥a
    • c ≥ a ≥ b
    • c ≥ b ≥ a
  • Case 1: a ≥ b ≥ c
    • (a @ b) = a, a @ c = a, b @ c = b
    • Hence (a @ b) @ c = a = a @ (b @ c)
    • Therefore the equality holds for the first case.
  • A complete proof requires that the equality be shown to hold for all 6 cases. But the proofs of the remaining cases are similar. Try them.

Without Loss of Generality

  • Example: Show that if x and y are integers  and both x∙y and x+y are even, then both x and y are even.
  • Proof: Use a proof by contraposition. Suppose  x and y are not both even. Then, one or both are odd. Without loss of generality, assume that x is odd. Then  x = 2m + 1 for some integer m.
    • Case 1: y is even. Then y = 2n for some integer n, so  x + y = (2m + 1) + 2n = 2(m + n) + 1 is odd.
    • Case 2: y is odd. Then y = 2n + 1 for some integer n, so x ∙ y = (2m + 1) (2n + 1) = 2(2m ∙ n +m + n) + 1 is odd.
  • We only cover the case where x is odd because the case where y is odd is  similar. The use phrase without  loss of generality (WLOG) indicates this. 

 

Existence Proofs

  • Proof of theorems of the form 
  • Constructive existence proof:
    • Find an explicit value of c, for which  P(c) is true.
    • Then   is   true by Existential Generalization (EG).
  • Example: Show that there is a positive integer that can be  written as the sum of cubes of positive integers in two different ways:
  • Proof:
    • 1729 is such a number since
    • 1729 = 10^3  + 9^3  = 12^3  + 1^3

 

Nonconstructive Proofs

  • In a nonconstructive existence proof, we assume no c exists which makes P(c) true and derive  a contradiction.
  • Example: Show that there exist irrational numbers x and y such that xy is rational.
  • Proof: We know that √2 is irrational (contradiction of above). Consider the number √2 √2 . If it is rational, we have two irrational numbers x and y with xy  rational, namely x = √2 and y = √2. But if √2 √2   is irrational, then we can let  x = √2 √2  and y = √2 so that
    • xy   = (√2 √2  )√2  = √2 (√2 √2)  = √2 2  = 2.

 

Counterexamples

 

Uniqueness Proofs

  • Some theorems asset the existence of a unique element with a particular property, ∃!x P(x). The two parts of a uniqueness proof are
    • Existence: We show that an element x with the property exists.
    • Uniqueness: We show that if y≠x, then y does not have the property.
  • Example: Show that if a and b are real numbers and  a ≠0, then there is a unique real number r  such that  ar + b = 0.
  • Solution:
    • Existence: The real number r = −b/a is a solution of ar + b = 0 because a(−b/a) + b = −b + b =0.
    • Uniqueness: Suppose that s is a real number such that   as + b = 0. Then ar + b = as + b, where r = −b/a.  Subtracting b from both sides and dividing by a shows that r = s.  

 

Proof Strategies for Conditionals – bring all the proving strategies together and figure out which one to use

  • Choose a method.
    • First try a direct method of proof.  
    • If this does not work, try an indirect method (e.g., try to prove the contrapositive).
  • For whichever method you are trying, choose a strategy.
    • First try forward reasoning.  Start with the axioms and known theorems and construct a sequence of steps that end in the conclusion.  Start with p and prove q, or start with ¬q and prove ¬p.
    • If this doesn’t work, try backward reasoning. When trying to prove q,  find a statement p that we can prove with the  property p → q.

 

Backward Reasoning

  • Example: Suppose that two people play a game taking turns removing, 1, 2, or 3 stones at a time from a pile that begins with 15 stones. The person who removes the last stone wins the game. Show that the first player can win the game no matter what the second player does.
    • Proof: Let n be the last step of the game.
      • Step n:    Player1 can win if the pile contains 1,2, or 3 stones.
      • Step n-1: Player2 will have to leave such a pile if the pile that he/she is faced with has 4 stones.
      • Step n-2: Player1  can leave 4 stones when there are 5,6, or 7 stones left at the beginning of his/her turn.
      • Step n-3: Player2  must leave  such a pile, if there are  8 stones .
      • Step n-4: Player1 has to have a pile with 9,10, or 11 stones to ensure that there are 8 left.
      • Step n-5: Player2  needs to be faced with  12 stones to be forced to leave 9,10, or 11.
      • Step n-6: Player1  can leave  12 stones by removing 3 stones.
      • Now reasoning forward, the first player can ensure a win by removing 3 stones and leaving 12.

 

Universally Quantified Assertions

  • Case 1. We show that if x is even then x2  is even using a direct proof (the only if part or necessity).
    • If x is even then x = 2k for some integer k.
    • Hence x2 =  4k2 = 2(2k2 ) which is even since it is an integer divisible by 2.
    • This completes the proof of case 1.
  • Case 2. We show that if x2 is even then x  must be  even (the if part or sufficiency). We use a proof by contraposition.
    • Assume x is  not even  and then show that 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 = 4k2 + 4k + 1 =  2(2k2 + 2k) + 1 which is odd and hence not even. This completes the proof of case 2.
    • Since x was arbitrary, the result follows by UG.
    • Therefore we have shown that x is even if and only if  x2 is even. 

 

 

Proving and Disproving

  • Example 1: Can we tile the standard checkerboard using dominos?
  • Solution: Yes! One example provides a constructive existence proof.

 

  • Example 2: Can we tile a checkerboard obtained by removing one of the four corner squares of a standard checkerboard?
  • Solution: 
    • Our checkerboard has 64 − 1 = 63 squares. 
    • Since each domino has two squares, a board with a tiling must have an even number of squares.
    • The number  63 is not even. 
    • We have a contradiction.

 

  • Example 3: Can we tile a board obtained by removing both the upper left and the lower right squares of a standard checkerboard? 
  • Solution:
    • There are 62 squares in this board.
    • To tile it we need 31 dominos. 
  • Key fact: Each domino covers one black and one white square. 
    • Therefore the tiling covers 31 black squares and 31 white squares.
    • Our board has either 30 black squares and 32 white squares or 32 black squares and 30 white squares.  
    • Contradiction!

 

Open Problems

  • Unsolved problems have motivated much work in mathematics. Fermat’s Last Theorem was conjectured more than 300 years ago. It has only recently been finally solved.
  • Fermat’s Last Theorem: The equation xn  + yn   = zn has no solutions in integers x, y, and z, with xyz≠0 whenever n is an integer with n > 2.
  • A proof was found by Andrew Wiles in the 1990s.

 

  • The 3x + 1 Conjecture: Let T be the transformation that sends an even integer x to x/2 and an odd integer x to 3x + 1. For all positive integers x, when we repeatedly apply the transformation T, we will eventually reach the integer 1.
    • For example, starting with x = 13:
      • T(13) = 3∙13 + 1 = 40, T(40) = 40/2  = 20, T(20) = 20/2  = 10,
      • T(10) = 10/2  = 5, T(5) = 3∙5 + 1 = 16,T(16) = 16/2  = 8,
      • T(8) = 8/2  = 4, T(4) = 4/2  = 2, T(2) = 2/2  = 1
    • The conjecture has been verified using computers up to    5.6 ∙ 1013  .

 

Other Proof Techniques

  • Later we will see many other proof methods:
    • Mathematical induction, which is a useful method for proving statements of the form ∀n P(n), where the domain consists of all positive integers.
    • Structural induction, which can be used to prove such results about recursively defined sets.
    • Cantor diagonalization is used to prove results about the size of infinite sets.
    • Combinatorial proofs use counting arguments.

 

 

 

 

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