Natural Deduction

Homework #1

~(~P & ~Q) P Q ~ (~ & Q) T F

~(~P & ~Q) P Q ~ (~ & Q) T F

~(~P & ~Q) P Q ~ (~ & Q) T F

~(~P & ~Q) P Q ~ (~ & Q) T F

~(~P & ~Q) P Q ~ (~ & Q) T F

~(~P & ~Q) P Q ~ (~ & Q) T F

~(~P & ~Q) P Q ~ (~ & Q) T F

(~(~P & ~Q) ↔ (P v Q)) So “~(~P & ~Q)” has the same truth-table as “(P v Q).” Why is that? Suppose I say: “you didn’t do your homework and you didn’t come to class on time.” When is this statement false? When either you did your homework or you came to class on time.

~(P → ~Q) P Q ~ (P → Q) T F

~(P → ~Q) P Q ~ (P → Q) T F

~(P → ~Q) P Q ~ (P → Q) T F

~(P → ~Q) P Q ~ (P → Q) T F

~(P → ~Q) P Q ~ (P → Q) T F

(~(P → ~Q) ↔ (P & Q)) So “~(P → ~Q)” has the same truth-table as “(P & Q).” Why is that? Suppose I say: “If you eat this spicy food, you will cry.” You might respond by saying “No, that’s not true: I will eat the spicy food and I will not cry.”

(~P v Q) P Q (~ v Q) T F

(~P v Q) P Q (~ v Q) T F

(~P v Q) P Q (~ v Q) T F

(~P v Q) P Q (~ v Q) T F

((~P v Q) ↔ (P → Q)) So (~P v Q) has the same truth-table as (P → Q). Why is that? Premise: Either I didn’t take my keys with me when I left home, or I lost them on the way to the office. Conclusion: Therefore, if I took them with me, I must have lost them on the way.

((~P v Q) ↔ (P → Q)) So (~P v Q) has the same truth-table as (P → Q). Why is that? Premise: If my mother remembers my birthday, I will get a letter in the mail. Conclusion: Therefore, either she won’t remember, or I’ll get a card.

(((P → Q) → P) → P) P Q (((P → Q) P) T F

(((P → Q) → P) → P) P Q (((P → Q) P) T F

(((P → Q) → P) → P) P Q (((P → Q) P) T F

(((P → Q) → P) → P) P Q (((P → Q) P) T F

(((P → Q) → P) → P) P Q (((P → Q) P) T F

So “(((P → Q) → P) → P)” is always true. Why is that So “(((P → Q) → P) → P)” is always true. Why is that? I’ll leave you to think about that. Later we’ll prove that it’s true.

(P & (~Q & R)) P Q R (P & (~ R)) T F

(P & (~Q & R)) P Q R (P & (~ R)) T F

(P & (~Q & R)) P Q R (P & (~ R)) T F

(P & (~Q & R)) P Q R (P & (~ R)) T F

(P & (~Q & R)) P Q R (P & (~ R)) T F

(P & (~Q & R)) P Q R (P & (~ R)) T F

Problem #2a P Q T F

With “→” P Q (Q → P) T F

With “v” P Q (~Q v P) T F

With “&” P Q ~(Q & ~P) T F

Problem #2b P Q T F

Obvious Solution P Q ~(P ↔ Q) T F

Problem #2c P Q T F

Opposite of #2a P Q ~(Q → P) T F

Another Good Solution P Q (~P & Q) T F

Problem #2d P Q T F

Any Tautology Will Do P Q (P v ~P) T F

Problem #2e P Q T F

You Should Have This Memorized P Q (P v Q) T F

Truth-tables and validity

The Truth-Table Test for Validity We know that an argument is deductively valid when we know that if it is true, then its conclusion must be true. We can use truth-tables to show that certain arguments are valid.

The Test Suppose we want to show that the following argument is valid: (P → Q) ~Q Therefore, ~P We begin by writing down all the possible truth-values for the sentence letters in the argument.

Write Down All the Possibilities Q T F

Write Truth-Table for Premises Q (P → Q) ~Q T F

Write Truth-Table for Conclusion P Q (P → Q) ~Q ~P T F

Look at Lines Where Premises are True Q (P → Q) ~Q ~P T *F F

derivations

The Parts of a Derivation 1 1. (A & D) A 2 2. (A → (B & C)) A 1 3. A 1 &E 1,2 4. (B & C) 2,3 →E 1,2 5. C 4 &E

Derived WFFs 1 1. (A & D) A 2 2. (A → (B & C)) A 1 3. A 1 &E 1,2 4. (B & C) 2,3 →E 1,2 5. C 4 &E

Derived WFFs A derivation proceeds in stages. Each stage involves writing a new WFF that we have proven. We number each of the stages (1, 2, …).

Rules of Inference 1 1. (A & D) A 2 2. (A → (B & C)) A 1 3. A 1 &E 1,2 4. (B & C) 2,3 →E 1,2 5. C 4 &E

Rules of Inference Whenever we write a new WFF that we have proven, we write the justification out to the right of the WFF. This justification includes: the name of the logical rule we used to prove the WFF, and the previous stages of the derivation that were needed by the rule (if any).

Dependencies 1 1. (A & D) A 2 2. (A → (B & C)) A 1 3. A 1 &E 1,2 4. (B & C) 2,3 →E 1,2 5. C 4 &E

Dependencies To the left of each stage we write the “dependencies.” So for example, 1,2 4. (B & C) 2,3 →E This line says that in stage #4 we proved the WFF “(B & C),” using the rule →E on lines 2 and 3. But we have only proved (B & C) assuming that 1 and 2 are true.

The Rule of Assumption: A Assumption is the easiest rule to learn. It says at any stage in the derivation, we may write down any WFF we want to. That WFF only depends on itself. For example, on line 57 we might write: 57 57. (((P ↔ (A & B)) → ~~~R) A

Turnstile The proper way to read this is “line 57 is provable from line 57” or “On the assumption that (((P ↔ (A & B)) → ~~~R) it can be proved that (((P ↔ (A & B)) → ~~~R).” We can re-write any line to state preciesely what we have proved. Line 57 would be: (((P↔(A & B))→~~~R) Ⱶ (((P↔(A & B))→~~~R)

&-Elimination: &E &E is also a very easy-to-learn rule. If we have proved (φ & ψ) on some line, then on any future line we may write down φ, and on any future line we may write down ψ. The result depends on everything (φ & ψ) depended on.

Example 1,2 4. (B & C) 2,3 →E 1,2 5. C 4 &E On Line 4 we have proved “(B & C)”

Example 1,2 4. (B & C) 2,3 →E 1,2 5. C 4 &E So on line 5 (or any future line) we can write down C.

Example 1,2 4. (B & C) 2,3 →E 1,2 5. C 4 &E Since the rule we used in arriving at 5 was &E applied to line 4, we write down “4 &E” as our justification.

Example 1,2 4. (B & C) 2,3 →E 1,2 5. C 4 &E Finally, the rule &E says the result (line 5) depends on everything the original conjunction depended on. Since line 4 depends on 1 and 2, we write “1,2” to the left of line 5.

Arrow Elimination: →E The →E rule says that if on one line we have a conditional (φ → ψ) and on another line we have the antecedent of the conditional φ then on any future line, we may write down the consequent of the conditional ψ depending on everything (φ → ψ) and φ depended on.

Example 2 2. (A → (B & C)) A 1 3. A 1 &E 1,2 4. (B & C) 2,3 →E On line 2 we have a conditional “(A → (B & C)),” and on line 3 we have its antecedent, “A.”

Example 2 2. (A → (B & C)) A 1 3. A 1 &E 1,2 4. (B & C) 2,3 →E So on line 4 we can write its consequent “(B&C).”

Example 2 2. (A → (B & C)) A 1 3. A 1 &E 1,2 4. (B & C) 2,3 →E Since the rule we used on line 4 was →E applied to lines 2 and 3, we write down “2,3 →E” to the right of 4.

Example 2 2. (A → (B & C)) A 1 3. A 1 &E 1,2 4. (B & C) 2,3 →E Finally, since line 2 depends on line 2, and line 3 depends on line 1, line 4 depends on lines 1 and 2. We copy those numbers to the left of 4.