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9:16 AM
Is there a model where $(\varphi \lor \psi), \neg \varphi \vdash \psi$ holds but $(\varphi \lor \psi), (\varphi \to \theta), (\psi \to \theta) \vdash \theta$ fails?
It would be intuitionistic.
 
 
4 hours later…
12:49 PM
To prove $\vdash (((A \to B) \to A) \to A$:
1. $(A \to B) \to A$ [assumption]
>2. $\neg A$ [assumption]
>>3. $A$ [assumption]
>>4. $\bot$
>>5. $B$ [explosion]
>6. $A \to B$ [$\to$intro 3-5]
>7. $A$ [MP 1 6]
>8. $\bot$
9. $\neg \neg A$ [$\neg$intro 2-8]
10. $A$ [$\neg$elim 9]
A1. $\varphi \to (\psi \to \varphi)$
A2. $(\varphi \to (\psi \to \chi)) \to ((\varphi \to \psi) \to (\varphi \to \chi))$

1. $\vdash \chi \to ((\chi \to \chi) \to \chi)$ A1
2. $\vdash (\chi \to ((\chi \to \chi) \to \chi)) \to ((\chi \to (\chi \to \chi)) \to (\chi \to \chi))$ A2
3. $\vdash (\chi \to (\chi \to \chi)) \to (\chi \to \chi)$ MP 1 2
4. $\vdash \chi \to (\chi \to \chi)$ A1
5. $\vdash \chi \to \chi$ MP 3 4
 
1:54 PM
@LeakyNun Do you mind next time typing in ASCII? I prefer not to have to read LaTeX (and I don't like the JS bookmarklet).
Anyway the answer is that the first one is in fact intuitionistically valid.
The second one is as well.
You kind of tricked me; I thought you somehow knew they weren't and went to think about them via Kripke semantics, only to find that the first was valid. Then I looked at the second and it was very obviously valid too.
@LeakyNun: It would be instructive for you to verify that they are valid using both the syntactic and semantic method, namely via explicit proof and via reasoning about all Kripke frames.
The deductive rules for intuitionistic logic are the same as in my Fitch-style system, except that ¬¬elim is replaced by the rule "⊥ |− A".
 
2:22 PM
@user21820 well, I can go from "(φ∨ψ),(φ→θ),(ψ→θ)⊢θ" to "(φ∨ψ),¬φ⊢ψ", but not the other way round, so I'm wondering if the second implies the first one as well.
 
Why do you want to go from one to the other? Just directly prove each one valid.
"(φ∨ψ),(φ→θ),(ψ→θ)⊢θ" is literally ∨elim.
It's hence the 'easier' one. "(φ∨ψ),¬φ⊢ψ" requires both ∨elim as well as the explosion rule "⊥ |− A".
 
have you read my proof that chi -> chi?
 
No I haven't; I don't find it fun to read Hilbert-style proofs.
Haha..
 
lol alright
@user21820 because somehow they're both named ∨elim
I don't really get the Kripke thing
 
@LeakyNun Oh I meant ∨elim in my system. So you're asking whether they are equivalent over the other rules? That's not the same as asking whether they are intuitionistically valid. I'll have to think again about it. Give me a while.
 
2:28 PM
@user21820 well if the second doens't imply the first then we can find models where the second is true and the first isn't, right?
 
@LeakyNun No because Kripke semantics is for models of intuitionistic logic, so to say that something is weaker than intuitionistic logic needs some other technique.
I'll explain Kripke frames later.
 
oh ok
∨elim: (a∨b),(a→c),(b→c)⊢c
 
Yea that's my version. I'm now trying to prove that you can obtain that using the other rule "(φ∨ψ),¬φ⊢ψ" over the rest of intuitionistic logic.
 
I'm just thinking out loud for now.
 
If I fail, then I'll believe that it can't be done. Heh.
 
2:37 PM
1. w⊩a or w⊩b
2. for all u≥w, u⊩a implies u⊩c
3. for all u≥w, u⊩b implies u⊩c
4. how not to use ∨elim to prove itself?
@user21820
 
@LeakyNun When you are in MS, you are always using classical logic.
 
@user21820 ok thanks
 
If your MS doesn't have classical logic, you practically can't do any reasoning about formal systems.
That's why you should find my earlier linked post about building blocks to be interesting, if you're interested in philosophy of mathematics.
Anyway I can't figure out how to prove the usual Or-Elim from the other one.
 
:o we have broken logic :P
The model in which a→(b∨c) ⊢ a→b ∨ a→c fails:
0: a
1a: a,b
1b: a,c
does this make any sense at all?
 
@LeakyNun That rule is valid in all Kripke frames, so your use of the word "model" does not make sense.
 
2:44 PM
how is that rule valid :o
you sent me a link earlier saying that it's invalid
 
Sorry I thought you were talking about the Or-Elim rule.
Lol.
Didn't read carefully enough.
 
¬¬a ⊢ a: can't construct a model :o
 
The intuition for Kripke frames is that atomic propositions are either known or unknown, and the frame is just a directed graph of worlds with no cycles.
 
I think 0: empty; 1:a should be a model
 
@LeakyNun Isn't this the example given at the end of Hanno's post?
 
2:47 PM
@user21820 I forgot it lol
yay I'm having intuition
 
@LeakyNun And isn't this the other example given by Hanno?
 
@user21820 this one is in fact given by Hanno
 
I think you just copied them, didn't you?
 
@user21820 paraphrased. not copied.
I'm trying to write down what I understand
should I ask a question about or-elim?
 
In any case, what you've written above do not sufficiently describe the Kripke frames. You must specify the ordering.
 
2:50 PM
@user21820 right, 0≤1a, 0≤1b, 0≤1.
0≤0, 1a≤1a, 1b≤1b, 1≤1.
 
@LeakyNun It would be an interesting question. I'm quite sure the answer is that the other Or-Elim is too weak...
 
well it's just between us now
the natural deductions are answered by either you or me or Mauro
 
My intuition is that the other Or-Elim requires too strong conditions. If it is the only rule available to eliminate disjunction, then it can't suffice for intuitionistic logic.
Because you can only use it when you have proven the negation of one of the disjuncts.
But the usual Or-Elim does not require you to have done that.
 
could you give me some simple sentence to model the counter-example of?
 
@LeakyNun You mean an example theorem that is intuitionistically valid but cannot be proven if you weaken the Or-Elim rule to the above variant?
 
2:58 PM
@user21820 no, I mean an example that's valid in classical but not intuition
(because I want to practise building Kripke frames)
 
@LeakyNun You have "not not A implies A". That's too simple for you? =)
 
@user21820 I just built it above and you said I copied Hanno, right?
 
Also "A or not A".
 
@user21820 same model :P
 
You could try De Morgan's.
I remember needing LEM/DNE to prove them in classical logic.
 
3:01 PM
one of them doesn't
 
Yea a couple of them don't.
 
I thought only one of the doesn't
¬(p∨q) ⊢ ¬p∧¬q
btw does a∨b ⊢ b∨a need ∨elim
 
Another one that doesn't: not A and not B implies not ( A or B )
But this one is surely non-intuitionistic: not ( not A and not B ) implies A or B.
Because the conditions don't provide enough information to construct the conclusion.
@LeakyNun Technically yes, unless you also add commutativity rules.
Here's another non-intuitionistic tautology: ( A implies B ) implies not A or B.
Which, if you recall, was something you said that started this whole discussion about intuitionistic logic.. Heheh..
Well but this one is too simple; it yields LEM and you already have a Kripke model against that.
Aww.. the other one also yields "not ( not A and not not A ) implies A or not A", and it's easy to prove "not ( not A and not not A )" intuitionistically, so again you get LEM.
@LeakyNun: Okay so it would be more interesting to have a tautology that does not imply LEM. The one in my linked question did, interestingly enough.
 
sorry, back now
@user21820 too many nots, lol
oh, I know one that doesn't include not
"⊢ (((A→B)→A)→A" is non-intuitionistic :P
@user21820 0:empty; 1:A,B?
 
3:23 PM
@LeakyNun Your frame doesn't satisfy ( ( A implies B ) implies A ).
 
@user21820 no that's for ( A implies B ) implies not A or B
click the little arrow on the reply icon
 
Oh lol okay.
But as I said we didn't need any frames.
It sufficed to get from those rules to LEM.
 
@user21820 you can't prove that you can't prove it without LEM
 
@LeakyNun You're making a logical error somewhere. If any of those rules are intuitionistically valid, then we can apply them in a suitable way to intuitionistically obtain LEM for every sentence. But we already know that LEM is not intuitionistically valid. Therefore all of those rules are not intuitionistically valid either.
 
nvm, I misread "get from those rules to LEM" as "get those rules from LEM"
 
3:27 PM
No problem.
@LeakyNun This one, on the other hand, can't be used to get LEM, I think. So it's a good other example.
 
how are the quantifiers realized in Kripke frames?
 
Kripke semantics (also known as relational semantics or frame semantics, and often confused with possible world semantics) is a formal semantics for non-classical logic systems created in the late 1950s and early 1960s by Saul Kripke and André Joyal. It was first conceived for modal logics, and later adapted to intuitionistic logic and other non-classical systems. The discovery of Kripke semantics was a breakthrough in the theory of non-classical logics, because the model theory of such logics was almost non-existent before Kripke (algebraic semantics existed, but were considered 'syntax in disguise...
 
I've looked at that page already
 
It seems that it doesn't treat ∃ as ¬∀¬, so last time the proof I gave didn't intuitionistically verify the original statement of the theorem.
 
0:
1: A,B (whence (A implies B) implies A)
 
3:29 PM
I vaguely recall reading somewhere that one could extend intuitionistic logic to allow deducing from ¬∀¬ to ∃.
 
but what does forall and exists mean there?
 
@LeakyNun As I said earlier this frame doesn't satisfy ( (A implies B) implies A ). What you want is a frame that satisfies it at some world but doesn't satisfy A at that world.
 
@user21820 how does that not satisfy that?
oops
 
World 1 satisfies it but also satisfies A. World 0 satisfies neither.
@LeakyNun It's just classical quantification over the domain of that one world.
 
@user21820 why is P∧Q(a) ⊢ P∧∀xQ(x) invalid in intuitionist?
: satisfy neither
A: satisfy both
B: satisfy neither
A,B: satisfy both
how can I build such frame...
 
3:37 PM
What is a?
 
free variable
 
You need a to be not occurring free in P as well.
Wait. No need.
But where is it stated that it's intuitionistically invalid?
 
4
Q: Natural Deduction First Order Logic ∃y∀x(P(x) ∨ Q(y))↔∀x∃y(P(x) ∨ Q(y))

Jesse GrootjenI'm working on some of my logic exercises for my end term exam in Predicate Logic. One of these exerecises is "Show with natural deduction that |- ∃y∀x(P(x) ∨ Q(y))↔∀x∃y(P(x) ∨ Q(y))" I'm getting the part that shows you can do ∃y∀x(P(x) ∨ Q(y)) |- ∀x∃y(P(x) ∨ Q(y)). It's the other way around i'm...

maybe I misunderstood
@KennyLau Your post is pure intuitionistic logic, but it only goes in 1 direction. The other direction is the hard one, it appears to be non constructive. — DanielV 17 hours ago
> For example, ¬¬ϕ≡(ϕ→⊥)→⊥ is known at w precisely if for any v∈W with w≤v there is z∈W with w≤z such that ϕ holds at z - in other words, you can never refute ϕ.
Could you explain why the interpretation of that statement is as stated?
 
I don't understand what you're asking.
 
why is "¬¬ϕ≡(ϕ→⊥)→⊥ known at w" equal to that condition?
 
3:53 PM
@LeakyNun This I can explain. ¬A is literally defined as (A→⊥) and is known at w iff you can't from w reach a world where A is known (since by definition of implication you can never know A. So ¬¬A is known at w iff from w you can never reach a world where (A→⊥) is known, which is equivalent to from w you cannot reach a world from which no reachable world knows A.
 
too meta
 
It's not.. Just think slowly.
 
>_>
 
From w you may be able to reach some world where A holds.
But all you need for ¬¬A to hold is that no matter where you go there is always a way to reach a world where A holds.
 
hmm.
wait until you see a triple negative :P
 
3:57 PM
If you can go to a world v where there is no way to reach a world where A holds, then we say that v refutes A.
 
what about the quantifier one?
 
That's the question I don't understand.
 
can we prove the backward direction, basically
 
Classically, you need the extra assumption that there is something.
As I told you before, you can never prove any existential sentence without having at least one axiom being an existential sentence.
So immediately you know you must use that quirky assumption.
 
yes, alright, we'll take that.
 
4:01 PM
Looks like you need LEM for Q(c), where c is something.
 
hmm
 
If Q(c) then done.
If not Q(c) then use the condition (right-hand side) to extract the required conclusion.
See it now?
 
alright
 
@LeakyNun: By the way, are you in london?
 
@user21820 soon
 
4:07 PM
Ah okay.
Oh wait.
Never mind.
 
school starts on 30
 
Okay.
I think you're very well prepared.
Lol.
 
lol
so, what's the solution to the (((A→B)→A)→A?
 
If you're bored you can try Manufactoria.
=P
 
thanks
 
4:12 PM
@LeakyNun I think 0→1 where 0 knows nothing and 1 knows A.
 
@user21820 hmm
oh
43 mins ago, by Leaky Nun
: satisfy neither
A: satisfy both
B: satisfy neither
A,B: satisfy both
this is wrong
: satisfy antecedent
A: satisfy both
B: satisfy neither
A,B: satisfy both
@user21820 any sentence for me?
 
@LeakyNun I don't know... Why don't you ask on Math SE for a propositional tautology that is not intuitionistically valid but is strictly weaker than the one in my linked question?
 
which linked question?
oh and you can't get LEM from (((A→B)→A)→A?
 
4:28 PM
@LeakyNun This one:
7
Q: Intuitionistic logic plus $A → B \lor C \vdash ( A → B ) \lor ( A → C )$

user21820The following is a classically valid deduction for any propositions $A,B,C$. $\def\imp{\rightarrow}$ $A \imp B \lor C \vdash ( A \imp B ) \lor ( A \imp C )$. But I'm quite sure it isn't intuitionistically valid, although I don't know how to prove it, which is my first question. If my conje...

@LeakyNun I suspect so.
Oops.
Wikipedia says it implies LEM.
 
@user21820 what's the name?
 
In logic, Peirce's law is named after the philosopher and logician Charles Sanders Peirce. It was taken as an axiom in his first axiomatisation of propositional logic. It can be thought of as the law of excluded middle written in a form that involves only one sort of connective, namely implication. In propositional calculus, Peirce's law says that ((P→Q)→P)→P. Written out, this means that P must be true if there is a proposition Q such that the truth of P follows from the truth of "if P then Q". In particular, when Q is taken to be a false formula, the law says that if P must be true whenever it...
 
oh
 
I should have tried proving LEM from it before guessing haha..
Anyway I got to go now.
See you next time!
 
bye
 
4:49 PM
@user21820 nice job tricking me into building finite automata :P
 
5:36 PM
13
A: Why Peirce's law implies law of excluded middle?

dtldarekIntuition: Peirce's law is of the form $$\big((A \to B) \to A\big) \quad \to \quad A,$$ that is, given $(A \to B) \to A$ we could deduce $A$. Therefore, we will try to construct $$((P \lor \neg P) \to \bot) \to (P \lor \neg P).$$ The trick is that for any $A \to B$ we can strengthen $A$ or relax...

 
 
2 hours later…
7:29 PM
⊢ A→B ∨ B→A: W={0→1,0→2}
0:
1: A
2: B
A→B is not known at 0 (1 is a counter-example)
B→A is not known at 0 (2 is a counter-example)
Proof that ¬¬¬A ⊢ ¬A:
1. ¬¬¬A premise
>2. A assumption
>>3. ¬A assumption
>>4. ⊥ !!! 2 3
>5. ¬¬A ¬intro 3-4
>6. ⊥ !!! 5 1
7. ¬A ¬intro 2-6
 
8:17 PM
@user21820 do you have any software/program/whatnot to write beautiful Fitch proofs?
Or Gentzen for that matter
 
 
1 hour later…
9:42 PM
0
A: Proof of $\exists x \exists y (\varphi(x)\rightarrow \psi(y)) \rightarrow \exists x (\varphi(x)\rightarrow \psi(x))$ in natural deduction

Kenny LauTo prove ∃x∃y[φ(x)→ψ(y)]→∃x[φ(x)→ψ(x)]: 01. ∃x∃y[φ(x)→ψ(y)] assumption 02. ∃y[φ(a)→ψ(y)] assumption, a 03. φ(a)→ψ(b) assumption, a, b 04. ¬φ(a) assumption, a, b 05. φ(a) assumption, a, b 06. ⊥ ...

 

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