I will reread your last reply tomorrow (when I will switch provability with demonstrating invalidity :)

There is one more question I'd like to ask, and I hope it is not difficult to you...

When there are two existential quantifiers stacked at the beginning of a premise, I first replace the first variable with a substituting instance (instead of Ex I assume "a"), and then the second variable (instead of Ey I assume "b"). I am not sure, however, what I am supposed to do when the existential quantifier I wish to substitute has a negation. For example: -(Ex)(Ey)(x≠y & (Fx & Fy)).


Should I substitute -(Ex) with negations only, e.g. (Ey)(-a ≠ y & (-Fa & Fy)) --and after think of a Sequent/Theorem Introduction (if there is one) that turns -a ≠ y to a=y--, or is there any other way to deal with quantifiers which contain negations?


Thank you!

Thank so much for you for your time. I will reread your replies in the weekend when I am off, with the hope that my mind will be sharper lol :)

This strategy of turning |≠(Ex)(Ay)((Rxy & ~Ryx) --> (Rxx <-> Ryy) into AxEy((Rxy & -Ryx) & -(Rxx <-> Ryy)) is mentioned by the author as well, but I am a little confused why turning (Ex)(Ay) to (Ax)(Ey) would somehow help to invalidate the first.

In ten days I have to take the exam and I still do not feel ready with everything....

Here is a conclusion that I am supposed to invalidate without any premises being given.

|≠ (Ex)(Ay)((Rxy & ~Ryx) --> (Rxx <-> Ryy)

I thought as a solution a domain with two members D= [a,b], where extension R=[<a,b>, <a,a>], hoping in this way to make the antecedent (Rxy&~Ryx) true and the consequent of the conditional false (since I am supposed to invalidate the sentence above). The author, however, does not give a solution to this problem. He just says that it was invented by Mendelson (?) and it can be solved only with four objects in the Domain (D=a,b,c,d or D=0,1,2,3?).

I am now wondering why the solution I gave with two objects will not do?

Regarding your solution... Do you mean identity elimination with IE on line 3? If yes, does IE allow us move b from the right of the first line to the left of the third line? I mean your proof looks good, but I need to know the rule that allows one change the place of b on line 3. Thanks!

1 (1) a=b Assumption
2 (2) a=c Assumption
1,2 (3) b=c 1,2 IE
2 (4) (a=b -> a=c) 1,2 ->I
(5) (a=b -> a=c) -> b=c 4,3 ->I
(6) Ax(x=b -> x=c)-> b=c 5, AI — jkg20

It helps a lot, although I have to familiarize with proving conditionals --and reread your previous replies. From what you say, I have a to find a way to support/prove the antecedent before I arrive at the consequent (something like Modus Ponens). The difficult part, which I need to familiarize with, is to support my antecedents through relying on the given premises or assumptions.

Thank you!

It seems I am not well acquainted with discharging assumptions, and that's the real issue here. I thought that I had discharged assumption of line 3 on line 4, where I apply existential elimination rule on lines 2 and 3... but you say that 3 is not discharged. I will google more about dischrging assumptions, but it would be helpful if you could tell me why line 4 is not considered a discharge of the assumption in line 3. Can you do that?

Thank you again for your time!!!

Ok I resolved the last exercise by myself, it turned to be easy.

Now there is one more conclusion. This is what I should prove (through using =E or =I rules):
|--nk (Ax)(x=b --> x=c)-->b=c

One way of proving this is...
1) (Ax)(x=b --> x=c)............. assumption
2) (a=b)--> (a=c)................ 1AE
3) a=b................................ assumption
4) a=c.................................2,3 -->E
5) b=a....... ........................3 COM (law of commutation)
6) b=c................................4,3 =E
7) (Ax) (x=b-->x=c)-->b=c

Does it look good?

I hope I am not boring you --you said you enjoyed memories from your undergraduate years (before you turned into a scientist I guess :), and I am trying all the available means in order not to fail again in the logic exam...

Professor said that I can avoid the dependency numbers (on the left) in my exam :) This is good news, because when you have many assumptions in a proof you have to take your time looking which ones are not discharged yet in order to enlist them on the left (with the other numbers where the current line depends). The are almost 20 questions to be answered in two hours, and the more time I save the better :) I may come up with a few more issues in the near future.

Everyone enjoy your weekend!

When I read his religious views, I came up with the impression that there cannot be a proof about the existence or non-existence of God because the universe in itself and the universe in our perception (scientific or not) cannot be the same thing. The universe in itself is an exact thing, but this exact thing (or the thing in itself) by no means is the same thing with the human perspective (scientific or not) of the universe. Since our human knowledge of the thing in itself is a kind of thesis or defense (viz. we can prove that there are reasons which justify our assertions about the universe, but we are far from claiming that we know the universe), it is impossible to say that this defense of this kind of knowledge can stand as a proof of whether God exist or does not. I gave up reading Kant's thoughts on religions and God long ago, in order to get some grasp of his ethics.... and hence I cannot be help in your research. But I guess you have to find out how Kant defines knowledge in order to understand all his metaphysics.

Ok, thank you so much for your time. You have been helpful, although I will reread your reply since the numbers on the left really confuse me. I will also ask my professor if I can avoid those numbers at the time of the exam. Unfortunately, I can't attend the logic class (since I can't leave work three times a week), and Forbes does not explain every detail in his book (I am a good reader, believe me).

By the way for proofs where I have to show that x=y, I suppose that x and y should pick the same referent. In other words, in a premise like (Ax)(Fx-->(Ey)(Gy & x=y)), I guess it is legitimate and helpful to substitute Fx with Fa and Gy with Ga (where both F and G predicates pick the same referent "a" and it becomes easy for me to show that x=y). Right?

Thank you again!

As I said, all help is appreciated for a beginner like me... and details are really important. Below I bring the second proof that gave me some hard time, since I am not sure what is the exact route of turning "=" into "≠". In the book I am given only the conclusion, no premises, and I try to solve it with some legitimate assumptions. I would greatly appreciate it if you or anyone else takes a little time with my solution to the exercise : Show |--NK (Ax)(Ay)((Fx & ~Fy)-->x≠y)

My proof:
1) Fa & ~Fb................................assumption
2) a=b..........................................assumption
3) Fa.................................. ........1&E
4) ~Fb................................ ........1&E
5) Fb............................................2,3 =E
6) ^^^contradiction^^^^.................. 4,5~E
7) ~(a=b)......................................2,6 ~I
8) (Fa & ~Fb)---> ~(a=b)....... ......1,7-->I
9) (Ay) ((Fa & ~Fy)--->a≠y).... ......8 AI -(as I said I am not sure how "=" turns into "≠" in a proof)
10) (Ax)(Ay)((Fx & ~Fy)---> x≠y)... 9 AI

jkg20

Insofar as everything depends on the details, I think you have been helpful already. My weakest point are assumptions (and I am confused with many braces and quantifiers as well). In the same book I am asked to show that |--NK (Ax)(Ay)((Fx & ~Fy)-->(x≠y) and again I am a little confused on whether (x≠y) can be equal to ~(x=y).

I guess in order to get such a conclusion I should start with: premise 1) Fa & ~Fb, premise 2) ~(a≠b). With the second premise I don't know what kind of proof or rules I may use (the chapter is referred to identity elimination rule =E), but I am not sure if premise 2 can be (a=b) --hoping to bring its opposite through introducing a negation (since I really do not know other way to turn = to ≠ in a proof).

Any ideas/suggestions on this last question? (I will try to solve it by myself also and hopefully will keep you updated).

Thank you for your support!

Nagase forgot this discussion :worry:

Anyone else willing to help?

Thanks!

Ok here is the solution I "found", without any sequent introductions... There seem to be many assumptions and I hope I have discharged them all.

1) (Ax)[(Ey)Tyx-->(Az)~Txz] premise
2) (Ax)[(Ey)Tyx-->Txx premise
3)(Ey)Tya-->Taa 2AE (viz. universal elimination of 2)
4) Tba-->Taa assumption
5)Tba assumption
6) Taa 4,5 -->E
7)(Ey)Tya-->(Az)~Taz 1AE
8) Tba-->(Az)~Taz assumption
9)Tba assumption (I assume it a second time since I am not sure if I should use line 5, already discharged, again)
10) (Az)~Taz 8,9 -->E
11) ~Taa 10AE
12) ^^^contradiction^^^ of 6 and 11
13) ~Tba 5,12~I
14) (Ay)~Tby 13AI
15) (Ax)(Ay)~Tby 14AI
16) (Ax)(Ay)~Tby 3,4,15 EE (viz. existential elimination)
17) (Ax)(Ay)~Tby 7,8,16 EE

By the way, these exercises do not help me pass the exam. I have to resolve them in order to make sense of everything and be able to interpret all the problems I will be given in the exam. Hence, I'd appreciate any corrections in my argument. Thank you again!

Ok, I appreciate your help. This is the example I am trying to resolve (from G. Forbes' book, p. 268, no answer is given though the author says there are 4 solutions/symbolizations).

With "Ax" are meant universal quantifiers, and with "Ex" existential quantifiers.

(Ax)[(Ey)Tyx--then-->(Az)~Txz], (Ax)[(Ey)Tyx--then-->Txx] |--NK (Ax)(Ay)~Txy

Thank you so much!!!