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?
@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".
@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.
@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.
@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.
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.
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.
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.
@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.
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...
@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.
I'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...
@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. — DanielV17 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?
@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.
@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?
The 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...
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...
Intuition:
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...
To 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. ⊥ ...