Question

Prove in a two column proof!
Given: (b^~c)-->a
~a
~c
~b --> d
Prove: d

Answer 1

Can you please clarify your quantors? I doubt that I have seen them before.
What is ^ ? and is:
\( \neg \text{ the same as } \sim\)

I guess that --> refers to a implication \(\implies \)

Answer 2

pardon me, quantors*, logical symbols in that manner.

Answer 3

^ means "and"

Answer 4

^ is and
~ is not
and --> is implies

Answer 5

Prove in a two column proof! Given:
1. (b^~c)-->a Premise
2. ~a Premise
3. ~c Premise
4. ~b --> d Premise

5. ~(b^~c ) 1,2 Modus Tollens
6. ~b v ~~ c 5
7. ~b v c 6 double negation

Answer 6

5. is De morgans

Answer 7

8. ~b 7 ~c
9. d 4,8

Answer 8

@kc_kennylau what are your reasons?

Answer 9

4 is disjunctive inference

Answer 10

sorry, I don't know, please ask @perl

Answer 11

8. ~b 3,7 Disjunctive syllogism
9. d 4,8 Modus Ponens

Answer 12

is 9 detachment

Answer 13

1. (b^~c)-->a Premise
2. ~a Premise
3. ~c Premise
4. ~b --> d Premise
5. ~(b^~c ) 1,2 Modus Tollens
6. ~b v ~~ c 5 Demorgan
7. ~b v c 6 double negation
8. ~b 3,7 Disjunctive syllogism
9. d 4,8 Modus Ponens

Answer 14

idk what modus tollens is

Answer 15

or pollens

Answer 16

its okay i think i got it

Answer 17

thanks!!!!

Answer 18

http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&ved=0CDsQFjAB&url=http%3A%2F%2Fhome.sbu.edu%2Frhughes%2Fsymbolic%2520logic%2F18%2520rules%2520of%2520deduction.doc&ei=JsHIUu7nCI3lsASquoCYDQ&usg=AFQjCNGcR5QywFkqYj3HBzvD5n5RsXpb-Q&sig2=M-OjM9sjD8AxsMtpW7VPlg&bvm=bv.58187178,d.cWc

Answer 19

@perl I think I know all the transformation rules, except their freaking names.

Answer 20

here, there are 18 rules of deduction

Answer 21

@perl how to know their freaking names?

Answer 22

well some of them are latin

Answer 23

Rule #1. Simplification (S): From p•q can be inferred p, and from p•q can be inferred q. From any conjunction of statements assumed true can be inferred the truth of any one of those statements taken separately.

Rule #2. Conjunction (CJ): From p and q can be inferred p•q. From any statements assumed true separately can be inferred the truth of the conjunction of those statements.

Rule #3. Modus Ponens (MP): From p->q and p can be inferred q. From a conditional statement and the affirmation of its antecedent can be inferred the affirmation of its consequent.

Rule #4. Modus Tollens (MT): From p->q and ~q can be inferred ~p. From a conditional statement and the negation of its consequent can be inferred the negation of its antecedent.

Rule #5. Disjunctive Syllogism (DS): From pvq and ~p can be inferred q, and from pvq and ~q can be inferred p. From a disjunction and the negation of a disjunct can be inferred the affirmation of the other disjunct. This rule functions as a process of elimination for various alternatives and works for both meanings of “either...or.”

Rule #6. Addition (AD): From p can be inferred pvq. From any statement assumed true can be inferred a disjunction composed of the original statement and any other statement whether true or false.

Rule #7. Hypothetical Syllogism (HS): From p->q and q->r can be inferred p->r. From any two conditional statements, where the consequent of one is identical to the antecedent of the other, can be inferred a conditional statement composed of the antecedent of the first and the consequent of the second.

Rule #8. Constructive Dilemma (CD): From p->q and r->s and pvr can be inferred qvs. From any two conditional statements and a disjunctive statement which affirms the antecedents of the two conditionals can be inferred a disjunction of the consequents of the two conditionals. This rule also works for the exclusive sense of “either...or.”

Answer 24

these are inference rules, then there are the rules of replacement

Answer 25

Rule #9. Double Negation (DN): p=~~p. Any statement is equivalent to its double negation.

Rule #10. Commutation (CM): (p•q)=(q•p) and (pvq)=(qvp). The order of statements in conjunctions and disjunctions in no way affects the truth value of the conjunction or disjunction (both exclusive and inclusive).

Rule #11. Association (AS): ((p•q)•r)=(p•(q•r)) and ((pvq)vr)=(pv(qvr)). The grouping of conjuncts or disjuncts in no way affects the truth value of conjunctions or disjunctions (both exclusive and inclusive).

Rule #12. Tautology (T): (p•p)=p and (pvp)=p. The repetition of statements in conjunctions or disjunctions is of no consequence logically.

Rule #13. Implication (I): (p->q)=(~pvq). A conditional statement is equivalent to a disjunction where the antecedent has been negated.

Rule #14. Contraposition (CP): (p->q)=(~q->~p). A conditional statement is equivalent to another conditional statement whose antecedent and consequent are reversed and negated.

Rule #15. Exportation (EX): ((p•q)->r)=(p->(q->r)). A conjunct, making up part of the antecedent of a conditional statement, can be exported to the consequent and connected to it as its antecedent.

Rule #16. Distribution (D): (p•(qvr))=((p•q)v(p•r)) and (pv(q•r))=((pvq)•(pvr)). A conjunct connected to another composed of disjuncts, can be connected to those disjuncts separately. A disjunct, connected to another composed of conjuncts, can be connected to those conjuncts separately.

Rule #17. Equivalence (EQ): (p<->q)=((p->q)•(q->p)) and (p<->q)=((p•q)v(~p•~q)). Two statements are materially equivalent if each can be inferred from the other. Also, two statements are materially equivalent if either both are true or both are false.

Rule #18. DeMorgan’s Theorem (DM): (p•q)=~(~pv~q) and (pvq)=~(~p•~q). Any conjunction is equivalent to a disjunction where the statement variables (taken separately) and the entire formula (taken as a whole) are negated. Any disjunction is equivalent to a conjunction where the statement variables and the entire formula have been negated.

Answer 26

rules of replacement are like 'substitution' rules. you can replace or substitute an expression for an equivalent expression, even parts of an expression can be changed.
inference rules are different, you have to use the whole line number