TheoremAny two distinct ones are orthogonal on .

*Proof:* We’ve already known that and are orthogonal, because

Let and be any two of these functions where ,

Since ,

But we also know and satisfy 2nd-order ODE

hence,

and

Since ,

TheoremAny two distinct ones are orthogonal on .

*Proof:* We’ve already known that and are orthogonal, because

Let and be any two of these functions where ,

Since ,

But we also know and satisfy 2nd-order ODE

hence,

and

Since ,

First, we need a partition of the sample space . Let’s say, . Bayes’ theorem is used when we know and for every , and we want to know for some . The word “partition” has two meanings, which is essential for the whole idea to work.

- s cover the whole sample space.
- s are disjoint.

Theorem 1 (Bayes’ theorem)

The formula seems a total mess at first glance, but if we look harder, we find out the numerator is due to the definition of conditional probability and the denominator is just due to the law of total probability. Taking one step closer, the numerator is one fraction of : relative to , and the denominator is the sum of all fractions of — itself: relative to . Hence the RHS of the formular is relative to — the conditional probability of given .

Theorem 2 (The Conditional Version of Bayes’ Theorem)

O.K. Here comes the trick. The whole idea behind Bayes’ theorem is that knowing every piece of ), and of course, to determine , kind of like switching the condition and the target event. So, Switch the two events around “” in the probability we want to determine, and write it down in the numerator. (We always multiply the probability of the conditional event to get something — ). Write (whatever on the right of “” you just wrote) in the numerator. Copy the numerator to the denominator, except change to, and sum it. ( is a specific number, and is every possible index of the partition.) And for conditional version of Bayes’ theorem, just plug “” before each “)” if there is no “” yet in the parenthesis:

It is important to notice that the numerator is one of the terms of the denominator. This is the main reason behind Example 2.3.6. Although the misdiagnosis rate is only 10% for those who actually has the disease, it’s also 10% for the healthy test taker to be misdiagnosed, a large part of the denominator, overwhelming the other part in the denominator as same as the numerator, which is relatively too high because the majority doesn’t has the disease. As a result, the conditional probability is exaggerated when taken for granted.

- Classify each of the following five expressions as either an atomic sentence, a negation, a conjunction, a disjunction, a conditional, a biconditional, or not an SC sentence. For each of the expressions that is an SC sentence, give a structure tree:
a) ¬(P ∨ (Q ∧ ¬R))

b) (¬P ∨ (Q ∧ ¬R))

c) ((¬P ∨ Q) ∧ ¬R)

d) ¬¬(¬¬φ ↔ ¬¬¬¬ψ)

e) (((A → B) → C) → ((A → C) → C))**Solution:**

a) negation

b) disjunction

c) conjunction

d) negation

e) conditional

- Give an example of each of the following:
a) a disjunction both of whose disjuncts are atomic.

b) a disjunction both of whose disjuncts are conjunctions both of whose

conjuncts are disjunctions.

c) a conditional whose antecedent is a biconditional and whose consequent is a

negated biconditional.**Solution:**

a) (P ∨ Q)

b) (((P ∨ Q) ∧ (P ∨ Q)) ∨ ((P ∨ Q) ∧ (P ∨ Q)))

c) ((P ↔ Q) → ¬(P ↔ Q)) - Describe an algorithm that, given an SC sentence as input, determines which of the six categories (atomic, negation, conjunction, disjunction, conditional, or biconditional) it falls into. Explain why your algorithm gives the correct classification of each of the SC sentences from problem 1.
**Solution:**def identify(s): assert type(s) is str if s[0] == "!": return s + " is a negation" if s[0] != "(": return s + " is a atomic sentence" p=0 i=1 while i < len(s): if s[i] == "(": p += 1 if s[i] == ")": p -= 1 if s[i] == "&" and p == 0: return s + " is a conjunction" if s[i] == "|" and p == 0: return s + " is a disjuction" if s[i] == "-" and p == 0: return s + " is a conditional" if s[i] == "+" and p == 0: return s + " is a conditional" if s[i] == "=" and p == 0: return s + " is a biconditional" if p < 0: return s + " is not a correct SC sentence" i += 1

- A substitution is an operation that replaces each atomic sentence by an SC sentence, making sure that the same replacement is made for all instances of a given atomic sentence. In other work, a substitution is a function
*s*associating SC sentences with SC sentences that meets the following conditions:s((φ ∨ ψ)) = (s(φ) ∨ s(ψ))

s((φ ∧ ψ)) = (s(φ) ∧ s(ψ))

s((φ → ψ)) = (s(φ) → s(ψ))

s((φ ↔ ψ)) = (s(φ) ↔ s(ψ))

s(¬φ) = ¬s(φ)Let

*s*be a substitution with*s*(“*P*“) = “(*P*∨*Q*),”*s*(“*Q*“) = “((*P*∨*Q*) →*R*),” and*s*(φ) = φ for φ an atomic sentence other than “*P*” or “*Q*“.a) What is

*s*(“(*P*∨*R*)”)?

b) What is*s*(“(*P*∨*Q*)”)?

c)What is*s*(*s*(“(*P*∨*R*)”))?

d) Give two distinct SC sentences φ and ψ with*s*(φ) =*s*(ψ).**Solution:**

a)*s*(“(*P*∨*R*)”)

= (*s*(“*P*“) ∨*s*(“*R*“))

= “((*P*∨*Q*) ∨*R*)”

b)*s*(“(*P*∨*Q*)”)

= (*s*(“*P*“) ∨*s*(“*Q*“))

= “((*P*∨*Q*) ∨ ((*P*∨*Q*) →*R*))”

c)*s*(*s*(“(*P*∨*R*)”))

=*s*(“((*P*∨*Q*) ∨*R*)”)

= (*s*(“(*P*∨*Q*)”) ∨*s*(“*R*“))

= “(((*P*∨*Q*) ∨ ((*P*∨*Q*) →*R*)) ∨*R*)”

d) φ = (*P*→*R*), ψ =*Q*

*s*(“*φ*“)

=*s*(“(*P*→*R*)”)

= (*s*(“*P*“) →*R*)

= “((*P*∨*Q*) →*R*)”

=*s*(“*ψ*“)

**1-5 pg. 27 #7**

If 12 balls are thrown at random into 20 boxes, what is the probability that no box will receive more than one ball?

Correct solution:

Erroneous solution:

# of different ways to arrange 12 balls and 19 separators, balls and separators are not distinguishable, unordered.

# of different ways to choose 12 boxes from 20 to put one ball for each, the order of elements in selection doesn’t matter.

Mistake: The Probability assigned to the sample space is *not equally probable*, hence the sample space is *not simple*. Situations that same boxes occupied by same number of balls but by different balls are counted only once, which is more likely to happen than all balls are put in a certain box, which is also counted once.

This mistake is just like another one: Tossing 3 coins simultaneously, defining the outcomes to be no heads, one heads, two heads, and three heads.

**1-6 pg. 27 #10**

A box contains 100 balls, of which r are red. Suppose that the balls are drawn from the box one at a time, at random, without replacement. Determine **(a)** the probability that the first ball drawn will be red; **(b)** the probability that the 50th ball drawn will be red; and **(c)** the probability that the last ball drawn will be red.

Solution:

A_{i} = {draw red at step i}

think of arranging the balls in 100 spots in a row.

**(a)**

**(b)**

sample space = sequences of length 50.

red on 50. There are 99 balls left, r choices to put red on 50.

, same as part a.

**(c)** As shown in part b, the particular draw doesn’t matter, probability is the same.

**1-8 pg.34 #16(b)**

The United States Senate contains two senators from each of the 50 states. What is the probability that a group of 50 senators selected at ramdom will contain on senator from each state?

Solution:

2 choices in selecting from the 1st states, 2 from the next…

or

100 choices in selecting the 1st person, 98 choices in selecting the next…and to eliminate the identical group but in different order when selected, divide by 50!.

actually

and

**1-11** If you throw r white balls into n boxes, r ≥ n, what is the number of distinguishable configurations in which no box remains empty?

Solution:

First of all, put 1 ball in each box from the beginning. r-n balls remain to be distributed in n boxes. (Use the ball/separator model)

However, figuring out the probability that no box remains empty is not that easy. Because if we use the ball/separator model to find the number of elements in sample space, the sample space would not be simple. And if we assume the balls are numbered, the outcomes in the event would be hard to determine. A solution involving independency and the probability of the union of finite number of events is available at pg. 63 – the collector’s problem.

**1-12** Given thirty people find the probability that among the twelve months there are six containing two birthdays and six containing three.

Solution:

30 people, 12 months.

Pr(6 months with 3 birthdays, 6 months with 2 birthdays)

Need to choose the 6 months with 3 or 2 birthdays, then the multinomial coefficient:

* Definition.*A set of sentences Ω is a

a) (φ ∧ ψ) ∈ Ω iff φ ∈ Ω and ψ ∈ Ω.

b) (φ ∨ ψ) ∈ Ω iff φ ∈ Ω or ψ ∈ Ω (or both).

c) (φ → ψ) ∈ Ω iff φ ∉ Ω or ψ ∈ Ω (or both).

d) (φ ↔ ψ) ∈ Ω iff φ and ψ are both in Ω or neither of them is.

e) ¬φ ∈ Ω iff φ ∉ Ω.

* Lemma.* A set of sentences Ω is a complete story iff it satisfies these two conditions:

1) Every finite subset of Ω is consistent.

2) For each sentence φ, either φ or ¬φ is an element of Ω.

* Lemma.* Suppose that every finite subset of ∆ is consistent. Take a sentence ψ. Then either every finite subset of ∆ ∪ {ψ} is consistent or every finite subset of ∆ ∪ {¬ψ} is consistent.

* Compactness Theorem.* A set Γ of SC sentences is consistent if and only if every finite subset of Γ is consistent.

* Corollary.* If φ is a logical consequence of a set of sentences ∆, then φ is a logical consequence of some finite subset of ∆.

* Soundness and Completeness Theorem.* A sentence χ is a logical consequence of a set of sentences Γ if and only if there is a derivation by which χ is derived from a premiss set that is included in Γ.

* Weak Completeness Theorem.* Every valid SC sentence is an SC theorem.

* Premiss Introduction Rule (PI).*You may write down any sentence you like if you take the sentence as its own premiss set.

* Conditional Proof Rule (CP).* If you have derived ψ with premiss set Γ, you may write (φ → ψ) with premiss set Γ ~ {φ}

* Modus Ponens Rule (MP).* If you have derived φ with premiss set Γ and (φ → ψ) with premiss set ∆, you may write ψ with premiss set Γ ∪ ∆.

* Modus Tollens Rule (MT).* If you have derived ψ with premiss set Γ and (¬φ → ¬ψ) with premiss set ∆, you may write φ with premiss set Γ ∪ ∆.

* Rule for Definition of Connectives (DC).* You may write an instance of any of the following six schemata with the empty premiss set:

((φ ∨ ψ) → (¬φ → ψ))

((¬φ → ψ) → (φ ∨ ψ))

((φ ∧ ψ) → ¬(φ → ¬ψ))

(¬(φ → ¬ψ) → (φ ∧ ψ))

((φ ↔ ψ) → ((φ → ψ) ∧ (ψ → φ)))

(((φ → ψ) ∧ (ψ → φ)) → (φ ↔ ψ))

* Theorem Substitution Derived Rule (TH).* If you have already proved φ from the empty set, you may, at any time in any derivation, write down any substitution instance of φ, again with the empty premiss set.

* Derived Rule for the Substitution of Equivalents (SE).* Suppose that φ has been derived from the premiss set Γ, that (χ ↔ θ) has been derived with premiss set ∆, and that ψ has been obtained from φ either by replacing χ with θ or by replacing θ with χ. Then you may derive ψ with premiss set Γ ∪ ∆.

* Derived Rule for Biconditional Introduction (BI).* If you have derived both (φ → ψ) and (ψ → φ), you may write (φ ↔ ψ), taking as premiss set the union of the premiss sets of the two conditionals.

TH1 | (¬¬P → P) | Double negation elimination |

TH2 | (Q → (P → Q)) | |

TH3 | ((P → Q) → ((Q → R) → (P → R))) | Principle of the syllogism |

TH4 | (((P → (Q → R)) → (Q → (P → R))) | |

TH5 | (P → ¬¬P) | Double negation introduction |

TH6 | (¬P → (P → Q)) | Law of Duns Scotus |

TH7 | ((¬P → P) → P) | Law of Clavius |

TH8 | (((P → Q) → R) → ((P → R) → R))) | |

TH9 | ((P → Q) → ((¬P → Q) → Q)) | |

TH10 | (P → (P ∨ Q)) | A disjunction introduction principle |

TH11 | (Q → (P ∨ Q)) | A disjunction introduction principle |

TH12 | ((P → R) → ((Q → R) → ((P ∨ Q) → R))) | Principle of disjunctive syllogism |

TH13 | ((P ∧ Q) → P) | A conjunction elimination principle |

TH14 | ((P ∧ Q) → Q) | A conjunction elimination principle |

TH15 | (P → (Q → (P ∧ Q))) | Conjunction introduction principle |

TH16 | ((P → Q) → ((Q → P) → (P ↔ Q))) | |

TH17 | ((P ↔ Q) → (P → Q)) | |

TH18 | ((P ↔ Q) → (Q → P)) | |

TH19 | ((P → Q) → (¬Q → ¬P)) | |

TH20 | ((¬Q → ¬P) → (P → Q)) | |

TH21 | ((P → Q) ↔ (¬Q → ¬P)) | Principle of contraposition |

TH22 | (¬(P ∧ Q) ↔ (¬P ∨ ¬Q)) | One of de Morgan’s laws |

TH23 | (P → P) | |

TH24 | (P ↔ P) | |

TH25 | (P ↔ ¬¬P) | |

TH26 | ((P ∨ Q) ↔ (¬P → Q)) | |

TH27 | ((P ∧ Q) ↔ ¬(P → ¬Q)) | |

TH28 | ((P ↔ Q) ↔ ¬(P → ¬Q)) | |

TH29 | ((P → (Q → R)) → ((P ∧ Q) → R)) | |

TH30 | (((P ∧ Q) → R) → (P → (Q → R))) | |

TH31 | ((P → (Q → R)) ↔ ((P ∧ Q) → R)) | |

TH32 | (¬(P ∨ Q) ↔ (¬ P ∧ ¬Q)) | One of de Morgan’s laws |

TH33 | ((P ∨ Q) ↔ (Q ∨ P)) | Commutative law for “∨” |

TH34 | ((P ∨ (Q ∨ R)) ↔ ((P ∨ Q) ∨ R)) | Associative law for “∨” |

TH35 | ((P ∨ P) → P) | |

TH36 | ((P ∨ P) ↔ P) | Idempotence for “∨” |

TH37 | ((P ∧ Q) ↔ (Q ∧ P)) | Commutative law for “∧” |

TH38 | ((P ∧ (Q ∧ R)) ↔ ((P ∧ Q) ∧ R)) | Associative law for “∧” |

TH39 | ((P ∧ P) ↔ P) | Idempotence for “∧” |

TH40 | ((P ∧ (Q ∨ R)) → ((P ∧ Q) ∨ (P ∧ R))) | |

TH41 | (((P ∧ Q) ∨ (P ∧ R)) → (P ∧ (Q ∨ R))) | |

TH42 | (((P ∧ (Q ∨ R)) ↔ ((P ∧ Q) ∨ (P ∧ R))) | Distributive law |

TH43 | ((P ∨ (Q ∧ R)) → ((P ∨ Q) ∧ (P ∨ R))) | Another distributive law |

Subject 24.241. Logic I. Homework due Thursday, September 29.

1.For this problem, let N be the set of natural numbers (nonnegative integers), let E be the set of even natural numbers, let P be the set of primes (integers>1 that aren’t divisible by any numbers other than themselves and one), and let S be the set of integers greater than 1 and less than 10.

a) List the members of S ～ P.

{4, 6, 8, 9}

b) List the members of S ～ (P ∪ E)

{9}

c) List the members of P ∩ E.

{2}

d) List the members of (S ～ P) ～ E.

{9}

e) List the members of S ～ (P ～ E).

{2, 4, 6, 8, 9}

2.For this problem, let I be {Mercury,Venus, Earth, Mars), and let C = {Spock,McCoy}.

a) List the functions from I to C, indicating which are one-one and which are onto.

There should be 2^{4}=16 functions. Due to pigeonhole theory, none of them is one-one and 14 of them are onto.

{<Mercury, Spock>, <Venus, Spock>, <Earth, Spock>, <Mars, Spock>}

{<Mercury, Spock>, <Venus, Spock>, <Earth, Spock>, <Mars, McCoy>} onto

{<Mercury, Spock>, <Venus, Spock>, <Earth, McCoy>, <Mars, Spock>} onto

{<Mercury, Spock>, <Venus, Spock>, <Earth, McCoy>, <Mars, McCoy>} onto

{<Mercury, Spock>, <Venus, McCoy>, <Earth, Spock>, <Mars, Spock>} onto

{<Mercury, Spock>, <Venus, McCoy>, <Earth, Spock>, <Mars, McCoy>} onto

{<Mercury, Spock>, <Venus, McCoy>, <Earth, McCoy>, <Mars, Spock>} onto

{<Mercury, Spock>, <Venus, McCoy>, <Earth, McCoy>, <Mars, McCoy>} onto

{<Mercury, McCoy>, <Venus, Spock>, <Earth, Spock>, <Mars, Spock>} onto

{<Mercury, McCoy>, <Venus, Spock>, <Earth, Spock>, <Mars, Spock>} onto

{<Mercury, McCoy>, <Venus, Spock>, <Earth, Spock>, <Mars, McCoy>} onto

{<Mercury, McCoy>, <Venus, Spock>, <Earth, McCoy>, <Mars, Spock>} onto

{<Mercury, McCoy>, <Venus, Spock>, <Earth, McCoy>, <Mars, McCoy>} onto

{<Mercury, McCoy>, <Venus, McCoy>, <Earth, Spock>, <Mars, Spock>} onto

{<Mercury, McCoy>, <Venus, McCoy>, <Earth, Spock>, <Mars, McCoy>} onto

{<Mercury, McCoy>, <Venus, McCoy>, <Earth, McCoy>, <Mars, Spock>} onto

{<Mercury, McCoy>, <Venus, McCoy>, <Earth, McCoy>, <Mars, McCoy>}

b) How many functions are there from I to I. of these how many are one-one? Onto? Both?

There are 4^{4}=256 functions.

4!=24 of them are both one-one and onto.

3.Use the method of truth table to identifl each of the following sentences as valid, inconsistent, or neither.

a) (((P ↔ (Q ↔ R)) ↔ ((P ↔ Q) ↔ ¬R)))

P |
Q |
R |
(((P |
↔ |
(Q |
↔ |
R)) |
↔ |
((P |
↔ |
Q) |
↔ |
¬ |
R))) |

1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 |

1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 |

1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 |

1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |

0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 |

0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 |

0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 |

0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 |

inconsistent

b) (((P → Q) → R) → ((P → R) → R))

P |
Q |
R |
(((P |
→ |
Q) |
→ |
R) |
→ |
((P |
→ |
R) |
→ |
R)) |

1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |

1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |

1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |

1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |

0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |

0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |

0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |

0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |

valid

c) (((P → (Q ∨ R)) ∨ ((P → Q) ∨ R))

P |
Q |
R |
(((P |
→ |
(Q |
∨ | R)) |
∨ | ((P |
→ |
Q) |
∨ | R)) |

1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |

1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 |

1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |

1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |

0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |

0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 |

0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 |

0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |

valid

4.For each of the following categories, either give an example or explain why there can’t be any example:

a) A tautological conditional whose antecedent is tautological.

((A ∨ ¬A) → (A ∨ ¬A))

b) An inconsistent conditional whose antecedent is inconsistent.

Can’t. Because if antecedent is inconsistent, the conditional is tautological.

c) A tautological disjunction neither of whose disjuncts are tautological.

A ∨ ¬A

d) A tautological conjunction neither of whose conjuncts are tautological.

Can’t. * SC Theorem 4. *A conjunction is tautological iff both its conjuncts are tautological.

e) An inconsistent sentence with no negation signs.

Can’t. Under the N.T.A. that every atomic sentences are true, every conjunction, disjunction, conditional and biconditional will also be true. Hence the sentence can’t be inconsistent without negation.

* Definition. * A

* Definition.* A truth assignment 𝕴 for a language for the sentential calculus is a

A conjunction is assigned the value 1 if and only if both its conjuncts are assigned 1.

A disjunction is assigned 1 if and only if one or both disjuncts are assigned 1.

A conditional is assigned 1 if and only if either its antecedent is assigned 0 or its consequent is assigned 1 (or both).

A biconditional is assigned 1 if and only if its components are both assigned the same value.

A negation is assigned 1 if and only if the negatum is assigned 1.

* Definition.* A sentence is said to be

* Definition.* An sentence is a

* SC Theorem 1. * An sentence is a tautology if and only if its negation is a contradiction.

* SC Theorem 2. * An SC sentence is contradictory iff its negation is a tautology.

* SC Theorem 3. * An SC sentence is indeterminate iff its negation is indeterminate.

* SC Theorem 4. * A conjunction is tautological iff both its conjuncts are tautological.

* SC Theorem 5. * For any SC sentences φ and ψ, if φ is a contradiction or ψ is a contradiction, then (φ ∧ ψ) is a contradiction.

* SC Theorem 6. * A conditional is contradictory iff its antecedent is a tautology and its consequent is a contradiction.

* Definition.* Two sentences are

* SC Theorem 7. * Two SC sentences φ and ψ are logically equivalent iff the biconditional (φ ↔ ψ) is a tautology.

* SC Theorem 8 (Augustus de Morgan). * ¬(φ ∨ ψ) is logically equivalent to (¬φ ∧ ¬ψ).

* SC Theorem 9 (also due to de Morgan). * ¬(φ ∧ ψ) is logically equivalent to (¬φ ∨ ¬ψ).

* SC Theorem 10. * Let 𝕴 be a N.T.A. and let φ, ψ, χ, and θ be SC sentences such that 𝕴(φ) = 𝕴(ψ) and such that χ and θ are just alike except that some occurrences of φ as a subsentence of χ have been replace by occurrences of ψ as subsentences of θ. The 𝕴(χ) = 𝕴(θ).

* SC Theorem 11. * If the SC sentences φ and ψ are logically equivalent and if χ and θ are just alike except that some occurrences of φ as a subsentence of χ have been replaced by occurrences of ψ as a subsentence of θ, then χ and θ are logically equivalent.

* Definition.* An SC sentence

* SC Theorem 12. * For any sentences φ and ψ, φ implies ψ iff the conditional (φ → ψ) is a tautology.

* SC Theorem 13. * A contradiction implies every sentence.

* SC Theorem 14. * A tautology is implied by every sentence.

* SC Theorem 15. * Two sentences are logically equivalent iff each implies the other.

* Definition.* An

* SC Theorem 16. * An SC argument

φ_{1
}φ_{2
}⋮

φ_{n
}∴ψ

is valid iff the conjunction (φ_{1} ∧ (φ_{2} ∧ … ∧ φ_{n})…) implies ψ.

* SC Theorem 17. * An SC argument

φ_{1}

φ_{2}

⋮

φ_{n}

∴ψ

is valid iff the conditional ((φ_{1} ∧ (φ_{2} ∧ … ∧ φ_{n})…) → ψ) is a tautology.

* Definition.* A sentence φ is a

* Definition.* A set of sentences is

* Simplest Example Lemma. * If there is an SC sentence with a given property, then there is a simplest SC sentence with the property. More precisely, there is an SC sentence having the property such that no proper subsentence has the property.

* Extension Theorem. * Any function assigning a numerical value, either 0 or 1, to every atomic sentence can be extended to a normal truth assignment in a unique way.

* No-Conflict Lemma. * If J and K are normal extensions of H and φ is in the domain of both of them, then J(φ) = K(φ).

* Definition.* A

s((φ ∨ ψ)) = (s(φ) ∨ s(ψ))

s((φ ∧ ψ)) = (s(φ) ∧ s(ψ))

s((φ → ψ)) = (s(φ) → s(ψ))

s((φ ↔ ψ)) = (s(φ) ↔ s(ψ))

s(¬φ)= ¬s(φ)

* Definition.* If φ is a sentence and s is a substitution, then s(φ) is said to be a

* Substitution Theorem 1. * Any substitution instance of a tautology is a tautology. Any substitution instance of a contradiction is a contradiction.

* Substitution Theorem 2. * Let s be a substitution. If φ implies ψ, then s(φ) implies s(ψ). If φ and ψ are logically equivalent, s(φ) and s(ψ) are logically equivalent. If φ is a logical consequence of Γ, then s(φ) is a logical consequence of {s(γ): γ ∈ Γ}.

* Substitution Theorem 3. * A sentence φ is consistent if and only if some substitution instance of φ is tautological.

* Substitution Theorem 4. * A sentence φ is tautological iff every substitution instance of φ is tautological iff every substitution instance of φ is consistent. A sentence ψ is contradictory iff every substitution instance of ψ is contradictory iff every substitution instance of ψ is invalid.

* Substitution Theorem 5. * Let φ and ψ be sentences whose only connectives are “∧,” “∨,” and “¬.” Then if φ implies ψ, ψ

根据gotosh的介绍，laptop已知有2个channel，一个负责primary monitor，一个负责2nd monitor

nvcap在显示kext里的info.plist中，大致结构为

04000000 0000xx00 xx000000 00000000 00000000

第一个xx为通道一，第二个为通道二

一般laptop每个通道有4个输出0000，其开关用bitmap的形式

测试数据

_____I E

01 00 XX CH2=0000 外显灭

01 04 XX CH2=0100

00 04 XX

01 02 XO CH2=0010 外显电亮 说明第2输出为外显

0F 0F XO 双通道全部接通，结果是，外显成为副显示器！！！内显仍旧灭。

02 00 XO 断开2通道，1通道外显接通，外显点亮，说明两通道都能各自控制4个输出

0D 02 XO CH1=1101 CH2=0010

FF 00 XO CH1全接通，CH2全开

测试过程中会遇到3种结果

1.外显亮内显灭 常态

2.外显灭内显亮 可以重新安装驱动 回到常态

3.内外显都灭 插入系统盘打开terminal，rm -r /volumes/disk0s2/system/library/extensions/nvinject.kext 回到状态2

只要有10.5环境，就无需刻盘

如果是GUID分区，直接进入原盘dmg，找到OSinstall.mpkg，自定义安装，去掉x11

如果是MBR分区，需要还原dmg至硬盘，用破解的OSinstaller替换才能安装

安装完毕用Uinstaller安装pc efi v9 chamelion 1.0.12，安装4个kext

用破解的boot替换10.5.6和SL的root下的boot

打DSDTpatch，生成DSDT.aml，放入sl的root

reboot，安装nvinject，网卡驱动

语言改成英文装声卡驱动

reboot后5国

安装voodoohda.kext 2.2

声卡成功驱动！！

迁移10.5.6，解决“关于本机”

至此，10.6完美运行！！！！2009.06.17

ci/qe开启必须条件：1.efi string为空; 2.IOProbeScore=0