• Rayan
    26
    So, here is the problem I must solve: This Block World only has three blocks: a, b, and c; and the on relation is axiomatised as follows: only on(a,b) and on(b,c) are true. The truth assignment is false for all the other atomic sentences.

    Now, the above relation is defined as follows:

    ∀x.∀z.(above(x,z) ⇔ on(x,z) ∨ ∃y.(above(x,y) ∧ above(y,z)))

    I am then asked whether the sentence above(c,a) is consistent with the sentences with on and above shown above. It seems to me that it is inconsistent, but when I check the right answers, it turns out that it is, in fact, consistent; but I can't really see how. Any help?

  • andrewk
    2.1k
    Are you saying that your axiom set is

    {<above axiom>, on(a,b), on(b,c)}

    where <above axiom> denotes the relation you wrote in the OP with the two forall quantifiers,
    or that it is

    {<above axiom>, on(a,b), on(b,c), ~on(b,a), ~on(c,b), ~on(a,c), ~on(c,a)}

    Or is the Block World a Model for the theory given by the axioms, such that the axioms are

    {on(a,b), on(b,c)}

    or

    {on(a,b), on(b,c), ~on(b,a), ~on(c,b), ~on(a,c), ~on(c,a)}

    and the relation in the OP is not an axiom but rather a feature of the model that is not entailed by the theory?

    The question of consistency depends on the answers to these questions. In particular, if ~on(c,a) is not provable then it is possible for on to mean 'to the left of' and the blocks to be arranged in a ring, so that every block is 'on' one other and every block is 'above' every other one.

    If you don't know what a Model is, don't worry, instead ask the question whether what you have described as a 'relation' in the OP is supposed to be an axiom. If not then what is it?
  • Rayan
    26
    Here is the link to the exercise: http://intrologic.stanford.edu/exercises/exercise_06_05.html
    That'll be better than me trying to explain it
  • Rayan
    26
    In particular, if ~on(c,a) is not provable then it is possible for on to mean 'to the left of' and the blocks to be arranged in a ringandrewk

    ~on(c,a) is part of the axiomatisation of on; therefore, on cannot mean to the left of.

    ask the question whether what you have described as a 'relation' in the OP is supposed to be an axiom.andrewk

    Yes, it is an axiom.
  • andrewk
    2.1k
    OK, with that information, we can conclude that it is consistent.

    One way to understand that is to consider the interpretation where above(x,y) is always true, for any x and y (eg 'above' means 'is a block'). Then the axiom defining 'above' is true for all x, y and z without having to use the 'on(x,z)' part of the disjunction. Hence there is no relationship at all between 'on' and 'above'.

    We could enforce the relationship by adding another item to the disjunction on the RHS of the axiom that defines 'above', viz:

    ∀x.∀z.(above(x,z) ⇔ ~on(z,x) ∧ (on(x,z) ∨ ∃y.(above(x,y) ∧ above(y,z))))

    With that revised axiom, above(c,a) makes the theory inconsistent because it enables us to deduce above(c,b), which then conflicts with the axiom on(b,c).

    The principle in cases like these is to remove all prior associations you have in your mind with the words that are used, because they will inject implicit assumptions into your analysis. If you find that hard to do, replace the words 'on' and 'above' by meaningless labels like 'P' and 'Q'. Once you've done that, consider various different truth tables ('interpretations') for P and Q that satisfy all the given axioms. If any of them make the statement you are testing true, then the statement does not make the system inconsistent.
  • Rayan
    26
    Thank you very much! Yes, I was injecting a lot of implicit assumptions.
  • tim wood
    8.7k
    (I'm assuming that the question is that if 1 below is true, then is there as assignment of -a, b, c that makes both 1 and 2 true.) The problem can be recast as:

    Given -a, b, c.

    1) (ab) ^ (bc)

    2 (Axz) k(xz) <=> (xz) or Ey: k(xy) ^ k(yz)

    2.5) because there is no a, 1) above is limited to just (bc).

    3) for xz=bc, k(xz) = k(bc) <=> (bc). True.

    4) then it becomes two questions: first, is there any other assignment of -a, b, c that permits the k function to yield (bc)? Ans., no.
    Second, is there any other assignment in k of -a, b, c that satisfies Ey: k(xy) ^ k(yz)? Per 1) and 2.5) only (bc) is available, so clearly there is no other assignment is available that satisfies both 1 an 2.

    Equally clearly I have probably not understood the problem. Will a kind soul set me straight?
  • Rayan
    26
    Hold on; there is something I actually didn't get. You said that we can interpret above(x,y) as always true, for any x and any y, and, then, you gave me the example that ‘‘above’’ might mean ‘‘is a block’’. But isn't above a binary relation constant? It can't be is a block. And I can't think of any interpretation under which above(x,y) is true for any x and any y.

    Edit: I got where I am wrong; I'm allowing implicit assumptions to infect my reasoning. I shouldn't think about it in terms of whether I can find concrete linguistic examples of relation constants which satisfy the sentence, but, rather, I should think in terms of truth tables; is that right?
  • Rayan
    26
    I have another question though; so, Γ is a set of Relational Logic sentences, and φ and ψ are individual Relational Logic sentences.
    Now, I am aksed whether this claim is true or false:

    If Γ ⊨ ¬φ[τ] for some ground term τ, then Γ ⊭ ∀x.φ[x]; true or false?

    Isn't it true? I mean, if Γ entails ¬φ for some given ground term, here τ, then it cannot be that Γ⊨ ∀x.φ[x], that is, it cannot be that Γ entails φ for all x; therefore, Γ ⊭ ∀x.φ[x].
    Am I missing something? Because when I check the answers, it turns out that it is false...
  • andrewk
    2.1k
    But isn't above a binary relation constant? It can't be is a block.Rayan
    Yes, that was a mistake. 'is a block' is unary. Instead let it be 'is the same shape as', which is binary.
  • andrewk
    2.1k
    If Γ ⊨ ¬φ[τ] for some ground term τ, then Γ ⊭ ∀x.φ[x]; true or false?

    Isn't it true?
    Rayan
    That sounds correct.
    Am I missing something? Because when I check the answers, it turns out that it is false...Rayan
    It sounds like you feel you've come up against something that is contrary to the above. I can't see it. What do you see that concerns you?
  • Rayan
    26
    I am following Stanford's introduction to propositional logic and at the end of each section they provide exercises and the answers to those exercises in order that you check whether your responses are correct.
    Well, when I click on “show answers,” it says that the claim “If Γ ⊨ ¬φ[τ] for some ground term τ, then Γ ⊭ ∀x.φ[x]” is false, whereas I think that it is true; and you seem to agree with me..
    Here is the link: http://intrologic.stanford.edu/exercises/exercise_06_07.html
  • andrewk
    2.1k
    Do you have a link? It could just be an error. Universities tend to have fairly high error rates in their materials these days, because all the fee income from undergrad students is funnelled into research to push the uni up the rankings list, rather than into quality control on student learning materials.
  • Rayan
    26
    Yes, I edited the last post to include the link
  • andrewkAccepted Answer
    2.1k
    It looks like an error to me. A set of truth assignments cannot satisfy both ¬φ[τ] and ∀x.φ[x] because the latter entails φ[τ].
  • Rayan
    26
    I know. God, I spent so much time fretting and thinking about what I might've possibly missed, all under the assumptions that the course material had to be correct.
    Anyway; I wanted to ask you something: how did you learn propositional logic? On your own or in university? Has it actually affected or improved your thinking? Very naive question, I know; but I'm a beginner, so I guess that that is a good excuse to ask naive questions
  • andrewk
    2.1k
    I am trained in mathematics, and use it in my work, but never studied formal logic until several years ago. I self-studied using what I could find on the internet, motivated by a desire to fully understand Godel's incompleteness theorems.

    I think it has helped my thinking a great deal. Whereas previously I could tell what was a valid proof, just by years of working on proofs in mathematics, I did not have the vocabulary and concepts to explain exactly what was wrong with a flawed proof, or to defend the correctness of a valid one. Studying logic gave me those tools. I have found it also very useful in debate, and in critically assessing tendentious statements by politicians or lobbyists. It equips one to spot the loaded question, the hidden assumption, and such like. Whenever somebody claims A therefore B, I find myself thinking 'what rule of inference did you use to get from A to B, and if you didn't use one, what's the basis for your assertion?'

    I am enthusiastic enough about it to even suggest that maybe some basic logic should be introduced to the curriculum in late High School.
  • Rayan
    26
    Interesting! Well, I asked because I don't really have any express, concrete end by which to justify my self-study of logic; I just enjoy learning it. I just got a bachelor degree in Biology and am switching to a double major in Literature/History, so I guess that I'm just very sporadic and eclectic; but sometimes I ask myself whether the time spent learning logic won't come at an opportunity-cost; and I wonder whether it will really yield any benefits that suit my future projects. But I can't know before I try, right?
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