If given :
1) it is not true that : 1>0 and 2+2=5
2) if ~1>0, then $\displaystyle 0\geq1$ ....... "~" :means not
3) if $\displaystyle 2+2\neq 5$,then 2+3=6
4) $\displaystyle 2+3\neq 6$
.......then prove: $\displaystyle 1\leq 0$
here's the idea, i leave it up to you to make it formal.
By (1), we have ~(1 > 0 and 2 + 2 = 5)
Apply DeMorgan's law, and we get ~(1 > 0) OR ~(2 + 2 = 5)
Assume ~(1 > 0)
Then, by (2) we have $\displaystyle 1 \le 0$. and the conclusion follows
Assume ~(2 + 2 = 5)
Then, by (3) we have 2 + 3 = 6
Since this contradicts (4), it must mean our assumption was wrong, and in fact, (2 + 2 = 5)
Since we have ~(1 > 0 and 2 + 2 = 5) and 2 + 2 = 5 is true, it must mean that we have ~(1 > 0) (this follows by applying a disjunctive syllogism, hopefully you see where to apply it). Since we have ~(1 > 0), the conclusion follows as it did in the first assumption
Hello, archidi!
The statements are:Given:
. . $\displaystyle (1)\;\text{It is not true that: }\underbrace{1\,>\,0}_p\text{ and }\underbrace{2+2\,=\,5}_q$
. . $\displaystyle (2)\:\text{If }\sim\underbrace{(1\,>\,0)}_{p}\text{, then }\,\underbrace{1\,\leq\,0}_{\sim p}$
. . $\displaystyle (3)\;\text{If }\underbrace{2+2\,\neq\,5}_{\sim q}\text{, then }\underbrace{2+3\,=\,6}_r$
. . $\displaystyle (4)\;\underbrace{2+3\,\neq\,6}_{\sim r}$
$\displaystyle \text{Then prove: }\:(5)\;\underbrace{1\,\leq\,0}_{\sim p}$
. . $\displaystyle \begin{array}{cc}(1) & \sim(p \wedge q) \\
(2) & \sim p \to \sim p \\ (3) & \sim q \to r \\ (4) & \sim r \\ \hline (5) & \sim p \end{array}$
(3) becomes: .$\displaystyle \sim r \to q$ . . . contrapositive of (3)
We are given: .$\displaystyle (4)\;\sim r$
Hence, we have: .$\displaystyle q\;\;{\color{blue}(a)}$ . . . Law of detachment
From (1), we have:
. . $\displaystyle \sim(p \wedge q) \;=\;\;\sim p\: \vee \sim q$ . . . DeMorgan's Law
. . . . . . . . $\displaystyle =\;\;\sim q \:\vee \sim p$ . . . Commutative
. . . . . . . . $\displaystyle = \;\;q \to \:\sim p\;\;{\color{blue}(b)}$ . . . Definition of Implication
We have: .$\displaystyle \begin{array}{cc}{\color{blue}(b)} & q \to\:\sim p \\ {\color{blue}(a)} & q\qquad \end{array}$
. . Therefore: .$\displaystyle \sim p$ . . . Law of Detachment
yes, we know that. but that does not matter here. as far as logic is concerned, truth is what our premises tell us it is. if the premise says 1 < 0, then we hold that to be "true".
this is a premise. it is true, regardless of whether it makes sense to you or not. premise (2) says the implication is true, so you must follow itHence ~(1>0) it does not logically imply $\displaystyle 1\leq 0$
yes, but it is given.The statement :~(1>0) -----> $\displaystyle 1\leq 0$ is just a conditional statement.
Like the statement : if 2+2=4 then Paris is in England
moreover, we could interpret ~p as $\displaystyle 1 \le 0$ since it is the only other option.
example, lets say there are only three possible movements of a particle, it can go up, it can stay still, or it can go down. it follows (by disjunctive syllogism) that if the the particle is not going up, then it must be staying still or going down. thus we can interpret ~(up) to mean (staying still or going down). you can write a proof for this.
similarly here, on the real line, we have ordering. a number is either greater than zero, equal to zero or less than zero. it follows that saying it is not greater than zero is the same as saying it is either equal to zero or less than zero, by the same reasoning of the preceding paragraph.
thus, with that in mind, Soroban's interpretation of ~p is ok.
however, if you want to use what is written and nothing more (no background knowledge of things like ordering and such) then follow my hint. otherwise, if you can get over this little hickup you have with Soroban's post, then use it. i do not see why your professor would not consider both methods correct
An argument ,$\displaystyle P\wedge Q\wedge R\wedge S\vdash T$ is valid
if the conditional ,$\displaystyle (P\wedge Q\wedge R\wedge S)\rightarrow T$ is a tautology irrespective of the values of P,Q,R,S,T
And in our case let:
P= ~(1>0and 2+2=5) ,AND since 1>0 is true but 2+2=5 is false then 1>0 and 2+2=5 is false and hence ~(1>0 and 2+2=5) is true.
Q= ~$\displaystyle (1>0)\rightarrow 1\leq 0$ is also true since: if false implies false is true
R= (2+2=/=5------>2+3=6) IS false since 2+2=/= 5 is true but 2+3=6 is false
ALSO S= $\displaystyle 2+3\neq 6$ is false,
Hence ,$\displaystyle P\wedge Q\wedge R\wedge S$ is false ,but
T=$\displaystyle 1\leq 0$ iS ALSO false and
Hence ,since false implying false is true ,the above argument is valid
SO by using the proper values of its constituents the argument can still be valid .
On the other hand if we take all the premises to be always true then ALL the arguments will be valid.
The answer by the professor will be given out in two days
you are not listening to what i am saying. in a logical argument, the premises we are given are statements that we assume to be true. whether or not they are actually true is a different matter. you can have a valid argument where none of the premises are true. and you can have an invalid argument where all the premises are true. please look up what the definition of "validity" and "soundness" is.
you were given an argument to prove its validity. that simply means you want to show that the conclusion follows logically from the premises. whether the premises are true or not is immaterial, your job is to show that IF they are true, the conclusion would have to be true.
this is not so.On the other hand if we take all the premises to be always true then ALL the arguments will be valid.
example:
1 + 1 = 2, therefore Jhevon likes heavy metal rock music.
clearly this argument is invalid, though the premise is true.
another:
1 < 2, therefore 2 = 3
and there are infinitely more silly arguments we can make with true premises.
again, look up the meaning of "valid" and "soundness"
O.k then lets take the following argument:
If:
1) if 2+2=5 ,then 2+3=6
2) 2+3=6.
Then 2+2=5
Now suppose .....2+2=5 is true ,2+3=6 is true and ,2+2=5-----> 2+3=6 also true .
Then the conclusion 2+2=5 is also true.
Is that argument valid??
But since we got a litle out of the discussion concerning Soroban"s solution,lets take the following argument:
given:
1) ~$\displaystyle (1>0)\rightarrow 1\leq 0$
2) ~(1>0)
Then to conclude $\displaystyle 1\leq 0$ ,we must use M.Ponens or detachment law.
Now the general form of that law is:
(p----->q and p)-------->q.
If however we put : p= 1>0 and ~p= $\displaystyle 1\leq 0$.
Like Soroban did in the solution of the argument of post #1
Then we will never be able to use M.Ponens to get the desired result
no, this argument is not valid. what you are saying here is [(P => Q) and Q] => P. this is false in general (it is only true if the converse of P => Q is true). you can see easily via truth table that this argument is invalid, but it is enough to know that P => Q does not mean that Q => P
furthermore, your intuition should tell you that the argument makes no sense. you made your conclusion the antecedent of one of the implications in your premises. you pretty much force yourself to beg the question in this case...which you did.
Soroban's argument is not as simple as you are making it seem here. watering down everything as you did here is bound to give rise to confusion. in fact, i am confused as to the point you are making here. you put a rule involving p and q and then you say it won't work because of the way we defined ~p. that makes no sense.But since we got a litle out of the discussion concerning Soroban"s solution,lets take the following argument:
given:
1) ~$\displaystyle (1>0)\rightarrow 1\leq 0$
2) ~(1>0)
Then to conclude $\displaystyle 1\leq 0$ ,we must use M.Ponens or detachment law.
Now the general form of that law is:
(p----->q and p)-------->q.
If however we put : p= 1>0 and ~p= $\displaystyle 1\leq 0$.
Like Soroban did in the solution of the argument of post #1
Then we will never be able to use M.Ponens to get the desired result
and i said this is not the case. you can have true premises but an invalid argument.