Mathematical Reasoning - Discrete Structures - Lecture Slides, Slides of Discrete Structures and Graph Theory

Download Mathematical Reasoning - Discrete Structures - Lecture Slides and more Slides Discrete Structures and Graph Theory in PDF only on Docsity!

Let’s proceed to…

• Mathematical Reasoning

Mathematical Reasoning

• We need mathematical reasoning to
• determine whether a mathematical argument is correct or incorrect and
• construct mathematical arguments.
• Mathematical reasoning is not only important for conducting proofs and program verification , but also for artificial intelligence systems (drawing inferences).

Terminology

• A lemma is a simple theorem used as an intermediate result in the proof of another theorem.
• A corollary is a proposition that follows directly from a theorem that has been proved.
• A conjecture is a statement whose truth value is unknown. Once it is proven, it becomes a theorem.

Rules of Inference

• Rules of inference provide the justification of the steps used in a proof.
• One important rule is called modus ponens or the law of detachment. It is based on the tautology (p∧(p→q)) → q. We write it in the following way:
• p
• p → q

• ---

• ∴ q

The two hypotheses p and p → q are written in a column, and the conclusion below a bar, where ∴ means “therefore”.

Rules of Inference

• p

• ---

• ∴ p∨q Addition

p∧q

---

∴ p Simplification

p q

---

∴ p∧q

Conjunction

¬q p→q

---

∴ ¬p

Modus tollens

p→q q→r

---

∴ p→r

Hypothetical syllogism

p∨q ¬p

---

∴ q

Disjunctive syllogism

Arguments

• Just like a rule of inference, an argument consists of one or more hypotheses and a conclusion.
• We say that an argument is valid , if whenever all its hypotheses are true, its conclusion is also true.
• However, if any hypothesis is false, even a valid argument can lead to an incorrect conclusion.

Arguments

• Which rule of inference was used in the last argument?
• p: “101 is divisible by 3.”
• q: “101^2 is divisible by 9.”

p p→q

---

∴ q

Modus ponens

Unfortunately, one of the hypotheses (p) is false. Therefore, the conclusion q is incorrect.

Arguments

• Another example:
• “If it rains today, then we will not have a barbeque today. If we do not have a barbeque today, then we will have a barbeque tomorrow. Therefore, if it rains today, then we will have a barbeque tomorrow.”
• This is a valid argument: If its hypotheses are true, then its conclusion is also true.

Arguments

• Another example:
• Gary is either intelligent or a good actor.
• If Gary is intelligent, then he can count from 1 to 10.
• Gary can only count from 1 to 2.
• Therefore, Gary is a good actor.
• i: “Gary is intelligent.”
• a: “Gary is a good actor.”
• c: “Gary can count from 1 to 10.”

Arguments

• i: “Gary is intelligent.” a: “Gary is a good actor.” c: “Gary can count from 1 to 10.”
• Step 1: ¬c Hypothesis
• Step 2: i → c Hypothesis
• Step 3: ¬i Modus tollens Steps 1 & 2
• Step 4: a ∨ i Hypothesis
• Step 5: a Disjunctive Syllogism Steps 3 & 4
• Conclusion: a (“Gary is a good actor.”)

Rules of Inference for Quantified Statements

• • (^) __________∀x P(x)
• ∴ P(c) if c∈U

Universal instantiation

P(c) for an arbitrary c___________________∈U

∴ ∀x P(x)

Universal generalization

______________________^ ∃x P(x) ∴ P(c) for some element c∈U

Existential instantiation

P(c) for some element c____________________∈U

∴ ∃x P(x)

Existential generalization

Rules of Inference for Quantified Statements

• Example:
• Every UMBC student is a genius.
• George is a UMBC student.
• Therefore, George is a genius.
• U(x): “x is a UMBC student.”
• G(x): “x is a genius.”

Proving Theorems

• Direct proof:
• An implication p→q can be proved by showing that if p is true, then q is also true.
• Example: Give a direct proof of the theorem “If n is odd, then n^2 is odd.”
• Idea: Assume that the hypothesis of this implication is true (n is odd). Then use rules of inference and known theorems to show that q must also be true (n^2 is odd).

Proving Theorems

• n is odd.
• Then n = 2k + 1, where k is an integer.
• Consequently, n^2 = (2k + 1) 2.
• = 4k 2 + 4k + 1
• = 2(2k 2 + 2k) + 1
• Since n^2 can be written in this form, it is odd.