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Comp280 hw02 solution

Module by: Ian Barland

Summary: Solutions for 2004.spring comp280 hw02.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

All (non-practice, non-extra-credit) problems due 2004.Feb.03 (Tue) 17:00.

Aside:

Rice Students: If you like, you can play WaterWorld on OwlNet, in /home/comp280/bin/waterworld. To run Waterworld on your home computer, download (from owlnet) the directory /home/scheme/plt/205/collects/waterworld/, and (from drscheme) add it as a teachpack.

Reasoning with Inference Rules

For proofs on this homework, Remember that each step must be justified by one of: "premise", a listed inference rule (.ps) (a previous line number and, if ambiguous, substitutions for the inference rule's metavariables), or a WaterWorld domain axiom.

Exercise 1

N.B.:

(Carried over from original hw01 assignment.)

(2 pts.) Fill in the blank reasons in the following proof that ∨∨ commutes, that is, χ ∨ υ ⊢ υ ∨ χχ ∨ υ ⊢ υ ∨ χ.

Table 1
1χ ∨ υχ ∨ υ  premise
2subproof:χ ⊢ υ ∨ χχ ⊢ υ ∨ χ  
2.a  χχ premise for subproof
2.b  υ ∨ χυ ∨ χ ∨∨Intro
3subproof:υ ⊢ υ ∨ χυ ⊢ υ ∨ χ  
3.a  υυ premise for subproof
3.b  υ ∨ χυ ∨ χ ____________
4υ ∨ χυ ∨ χ  ____________

Solution

  • Line 3b justification: ∨∨Intro, by line 3a.
  • Line 4 justification: ∨∨Elim, by lines 1,2,3, where φφ=χχ and ψψ=υυ and θθ=υ ∨ χυ ∨ χ

Exercise 2

(5pts.) Show that τ ∧ χτ → υχ → ω ⊢ υ ∧ ωτ ∧ χτ → υχ → ω ⊢ υ ∧ ω. (It may take around 8 steps.)

Solution

This is basically just two instances of →→Elim, plus a little manipulation of ∧∧.

Table 2
1τ ∧ χτ ∧ χ premise
2ττ ∧∧Elim (left), by line 1, where φφ=ττ and ψψ=χχ
3τ → υτ → υ premise
4υυ →→Elim, by lines 2,3, where φφ=ττ and ψψ=υυ
5χχ ∧∧Elim (right), by line 1, where φφ=ττ and ψψ=χχ
6χ → ωχ → ω premise
7ωω →→Elim, by lines 5,6, where φφ=χχ and ψψ=ωω
8υ ∧ ωυ ∧ ω ∧∧Intro, by lines 4,7, where φφ=υυ and ψψ=ωω

Exercise 3

(4pts) Using the inference rule RAA, prove ¬χ ⊢ ¬χ ∧ υ¬χ ⊢ ¬χ ∧ υ.

Solution

We observe that to use RAA to conclude ¬χ ∧ υ¬χ ∧ υ will require a subproof of ¬¬χ ∧ υ ⊢ false¬¬χ ∧ υ ⊢ . (That is, the rule's φφ refers to a WFF which is already a negation: "¬χ ∧ υ¬χ ∧ υ". Not a problem at all.)

Table 3
1¬χ¬χ  premise
2subproof:¬¬χ ∧ υ ⊢ false¬¬χ ∧ υ ⊢   
2.a  ¬¬χ ∧ υ¬¬χ ∧ υ premise for subproof
2.b  χ ∧ υχ ∧ υ ¬¬Elim, by line 2.1
2.c  χχ ∧∧Elim, by line 2.2
2.d  false falseIntro, by lines 1,2.3
3¬χ ∧ υ¬χ ∧ υ  RAA, by line 2, where φφ=¬χ ∧ υ¬χ ∧ υ

Exercise 4

(6pts.) Show that ¬W-safe ∨ ¬Y-unsafe ⊢ W-unsafe ∨ Y-safe¬W-safe ∨ ¬Y-unsafe ⊢ W-unsafe ∨ Y-safe.

hint:

Use the ∨∨Elim inference rule to get the final result.

Solution

Table 4
1¬W-safe ∨ ¬Y-unsafe¬W-safe ∨ ¬Y-unsafe  premise
2subproof:¬W-safe ⊢ W-unsafe ∨ Y-safe¬W-safe ⊢ W-unsafe ∨ Y-safe  
2.a  ¬W-safe¬W-safe premise for subproof
2.b  ¬W-safe → W-unsafe¬W-safe → W-unsafeWaterWorld domain axiom
2.c  W-unsafeW-unsafe →→Elim
2.d  W-unsafe ∨ Y-safeW-unsafe ∨ Y-safe ∨∨Intro
3subproof:¬Y-unsafe ⊢ W-unsafe ∨ Y-safe¬Y-unsafe ⊢ W-unsafe ∨ Y-safe  
3.a  ¬Y-unsafe¬Y-unsafe premise for subproof
3.b  ¬Y-unsafe → Y-safe¬Y-unsafe → Y-safe WaterWorld domain axiom
3.c  Y-safeY-safe →→Elim
3.d  W-unsafe ∨ Y-safeW-unsafe ∨ Y-safe ∨∨Intro
4W-unsafe ∨ Y-safeW-unsafe ∨ Y-safe  ∨∨Elim

Exercise 5

(5pts) For the following WaterWorld proof of X-has-1 ⊢ W-unsafe ∨ Y-unsafeX-has-1 ⊢ W-unsafe ∨ Y-unsafe, provide the missing steps and/or reasons:

Table 5
1X-has-1X-has-1  ____________
2____________  WaterWorld domain axiom
3____________  →→Elim
4subproof:W-safe ∧ Y-unsafe ⊢ W-unsafe ∨ Y-unsafeW-safe ∧ Y-unsafe ⊢ W-unsafe ∨ Y-unsafe  
4.a  W-safe ∧ Y-unsafeW-safe ∧ Y-unsafe premise for subproof
4.b  Y-unsafeY-unsafe____________
4.c  W-unsafe ∨ Y-unsafeW-unsafe ∨ Y-unsafe ____________
5subproof:¬W-safe ∧ Y-unsafe ⊢ W-unsafe ∨ Y-unsafe¬W-safe ∧ Y-unsafe ⊢ W-unsafe ∨ Y-unsafe  
5.a  ¬W-safe ∧ Y-unsafe¬W-safe ∧ Y-unsafe premise for subproof
5.b  W-unsafe ∧ Y-safeW-unsafe ∧ Y-safe CaseElim (left), where φφ=____________, and ψψ=____________.
5.c  ____________ ____________
5.d  W-unsafe ∨ Y-unsafeW-unsafe ∨ Y-unsafe ____________
6W-safe ∧ Y-unsafe ∨ ¬W-safe ∧ Y-unsafeW-safe ∧ Y-unsafe ∨ ¬W-safe ∧ Y-unsafe From class, φ ∨ ¬φφ ∨ ¬φ, with φ____________φ____________
7W-unsafe ∨ Y-unsafeW-unsafe ∨ Y-unsafe  ____________

Solution

Since Step 4 is of the form φ → θφ → θ, and Step 5 is of the form ¬φ → θ¬φ → θ, this is enough to prove θθ by ∨∨Elim, once we include (from class) φ ∨ ¬φφ ∨ ¬φ.

Table 6
1X-has-1X-has-1  premise
2X-has-1 → W-safe ∧ Y-unsafe ∨ W-unsafe ∧ Y-safeX-has-1 → W-safe ∧ Y-unsafe ∨ W-unsafe ∧ Y-safe  WaterWorld domain axiom
3W-safe ∧ Y-unsafe ∨ W-unsafe ∧ Y-safeW-safe ∧ Y-unsafe ∨ W-unsafe ∧ Y-safe  →→Elim
4subproof:W-safe ∧ Y-unsafe ⊢ W-unsafe ∨ Y-unsafeW-safe ∧ Y-unsafe ⊢ W-unsafe ∨ Y-unsafe  
4.a  W-safe ∧ Y-unsafeW-safe ∧ Y-unsafe premise for subproof
4.b  Y-unsafeY-unsafe ∧∧Elim
4.c  W-unsafe ∨ Y-unsafeW-unsafe ∨ Y-unsafe ∨∨Intro
5subproof:¬W-safe ∧ Y-unsafe ⊢ W-unsafe ∨ Y-unsafe¬W-safe ∧ Y-unsafe ⊢ W-unsafe ∨ Y-unsafe  
5.a  ¬W-safe ∧ Y-unsafe¬W-safe ∧ Y-unsafe premise for subproof
5.b  W-unsafe ∧ Y-safeW-unsafe ∧ Y-safe CaseElim (left), where φφ=W-safe ∧ Y-unsafeW-safe ∧ Y-unsafe, and ψψ=W-unsafe ∧ Y-safeW-unsafe ∧ Y-safe
5.c  W-unsafeW-unsafe ∧∧Elim
5.d  W-unsafe ∨ Y-unsafeW-unsafe ∨ Y-unsafe ∨∨Intro
6W-safe ∧ Y-unsafe ∨ ¬W-safe ∧ Y-unsafeW-safe ∧ Y-unsafe ∨ ¬W-safe ∧ Y-unsafe From class, φ ∨ ¬φφ ∨ ¬φ, with φW-safe ∧ Y-unsafeφW-safe ∧ Y-unsafe
7W-unsafe ∨ Y-unsafeW-unsafe ∨ Y-unsafe  ∨∨Elim, by lines 5,6, with φW-safe ∧ Y-unsafe ψ¬W-safe ∧ Y-unsafe θW-unsafe ∨ Y-unsafeφW-safe ∧ Y-unsafeψ¬W-safe ∧ Y-unsafeθW-unsafe ∨ Y-unsafe

Exercise 6

[practice; 0pts]

Figure 1: A sample WaterWorld board
Figure 1 (hwA-1.png)

Given the above figure, and using any of the immediately obvious facts as premises, prove that location P is safe by using our proof system and the WaterWorld domain axioms.

While this proof is longer (nearly 30 steps), it's not too bad. To make life easier, you may also use the following:

  • ¬φ ⊢ ¬φ ∧ ψ¬φ ⊢ ¬φ ∧ ψ
  • Q-has-1 → P-safe ∧ R-safe ∨ P-safe ∧ W-safe ∨ R-safe ∧ W-safeQ-has-1 → P-safe ∧ R-safe ∨ P-safe ∧ W-safe ∨ R-safe ∧ W-safe.
  • Not only is Y-safeY-safe an allowable premise (reading off the given board), but so is ¬Y-unsafe¬Y-unsafe.

Solution

Table 7
1Q-has-1Q-has-1  premise
2X-has-1X-has-1  premise
3¬Y-unsafe¬Y-unsafe  premise
4X-has-1 → Y-unsafe ∨ W-unsafeX-has-1 → Y-unsafe ∨ W-unsafe th'm: previous problem (and, →→Intro)
5W-unsafe ∨ Y-unsafeW-unsafe ∨ Y-unsafe  →→Elim, by lines 2,4
6Y-unsafe ∨ W-unsafeY-unsafe ∨ W-unsafe  ∨∨ commutes; line 5
7W-unsafeW-unsafe  CaseElim, by lines 3,6
8subproof:¬¬P-safe ∧ W-safe ⊢ false¬¬P-safe ∧ W-safe ⊢   
8.a  ¬¬P-safe ∧ W-safe¬¬P-safe ∧ W-safe premise for subproof
8.b  P-safe ∧ W-safeP-safe ∧ W-safe ¬¬Elim, by line 8.1
8.c  W-safeW-safe ∧∧Elim
8.d  W-safe → ¬W-unsafeW-safe → ¬W-unsafe WaterWorld domain axiom
8.e  ¬W-unsafe¬W-unsafe →→Elim, by lines 8.2,8.3
8.f  false falseIntro, by lines 7,8.4
9¬P-safe ∧ W-safe¬P-safe ∧ W-safe  RAA, by line 8
10subproof:¬¬R-safe ∧ W-safe ⊢ false¬¬R-safe ∧ W-safe ⊢   
10.a  ¬¬R-safe ∧ W-safe¬¬R-safe ∧ W-safe premise for subproof
10.b  R-safe ∧ W-safeR-safe ∧ W-safe ¬¬Elim, by line 10.1
10.c  W-safeW-safe ∧∧Elim
10.d  W-safe → ¬W-unsafeW-safe → ¬W-unsafe WaterWorld domain axiom
10.e  ¬W-unsafe¬W-unsafe →→Elim, by lines 10.2,10.3
10.f  false falseIntro, by lines 7,10.4
11¬R-safe ∧ W-safe¬R-safe ∧ W-safe  RAA, by line 10
12Q-has-1 → P-safe ∧ R-safe ∨ P-safe ∧ W-safe ∨ R-safe ∧ W-safeQ-has-1 → P-safe ∧ R-safe ∨ P-safe ∧ W-safe ∨ R-safe ∧ W-safe  Allowed by problem statement
13P-safe ∧ R-safe ∨ P-safe ∧ W-safe ∨ R-safe ∧ W-safeP-safe ∧ R-safe ∨ P-safe ∧ W-safe ∨ R-safe ∧ W-safe  →→Elim, by lines 1,12
14R-safe ∧ W-safe ∨ P-safe ∧ R-safe ∨ P-safe ∧ W-safeR-safe ∧ W-safe ∨ P-safe ∧ R-safe ∨ P-safe ∧ W-safe  ∨∨ commutes (ex.1), by line 13
15P-safe ∧ R-safe ∨ P-safe ∧ W-safeP-safe ∧ R-safe ∨ P-safe ∧ W-safe  CaseElim, by lines 9,14
16P-safe ∧ W-safe ∨ P-safe ∧ R-safeP-safe ∧ W-safe ∨ P-safe ∧ R-safe  Theorem: ∨∨ commutes (ex.1), line 15.
17P-safe ∧ R-safeP-safe ∧ R-safe  CaseElim, by lines 11,16.
18P-safeP-safe ∧∧Elim -- Whew!

The subproofs could easily have been pulled out into lemmas (making the proof clearer, exactly as subroutines make programs clearer, even if only called once). Observe how the second subproof had to repeat lines identical from the first -- it would have been incorrect to cite those earlier subproof lines, because the previous subproof had been completed.

note:
Interestingly, we didn't need to use R-safeR-safe as a premise. (In fact, we nearly proved that ¬R-safe¬R-safe would have been inconsistent with the other premises.)

Exercise 7

(6 pts)

  • State where on a board pirates could be positioned, so that: P-has-1 ∧ U-has-1 ∧ W-has-1P-has-1 ∧ U-has-1 ∧ W-has-1, but X-safeX-safe.
  • Compare this with a previous theorem (example 11) B-has-1 ∧ G-has-1 ∧ J-has-1 → K-unsafeB-has-1 ∧ G-has-1 ∧ J-has-1 → K-unsafe, the same idea shifted down a couple of rows. Suppose we try to translate this proof so as to conclude ¬X-safe¬X-safe (clearly untrue, by the above). What is the first step of the proof which doesn't hold when B,G,J,K are replaced with P,U,W,X, respectively? (Just give a line number; no explanation needed. Your answer will be of the form "Lemma A line 1" or "main proof line 2".)
  • If you replace such steps with the closest true formulas you can ("translating" instead of "transliterating", if you will), 1 where is the first line of the proof which fails and cannot be patched? (Just give a line number; no explanation needed. Your answer will be of the form "Lemma A line 1" or "main proof line 222".)

Solution

You could have pirates at H and V, (and, no pirates at X, N, or Q), and still have P-has-1 ∧ U-has-1 ∧ W-has-1P-has-1 ∧ U-has-1 ∧ W-has-1, be true on the given board.

Lemma B, line 5: After a direct substition, the formula we obtain isn't an actual domain axiom any more. The domain axiom B1 had as its conclusion that there is exactly one pirate in A and C, while the domain axiom P1 is not exactly analogous: its conclusion is exactly one pirate in N,Q,H. So a translation should really use the correct domain axiom P1.

Lemma B, line 8: By patching line 5 we can still continue to step 6, inferring a disjunction with 3 terms, still using →→Elim (namely, H-safe ∧ N-safe ∧ Q-unsafe ∨ H-safe ∧ N-unsafe ∧ Q-safe ∨ H-unsafe ∧ N-safe ∧ Q-safeH-safe ∧ N-safe ∧ Q-unsafe ∨ H-safe ∧ N-unsafe ∧ Q-safe ∨ H-unsafe ∧ N-safe ∧ Q-safe). The fundamental problem comes up in line 8: we can't apply the cited theorem (φ ∨ ψ ⊢ ¬φφ ∨ ψ ⊢ ¬φ). There's no obvious way to patch it to still end up with a single conjunction like N-safe ∧ H-safe ∧ Q-unsafeN-safe ∧ H-safe ∧ Q-unsafe. We wonder whether, if we're clever enough and somehow twiddle ∧∧'s associativity or we extend the theorem to handle disjunctions with three terms, if we can get the desired result? No -- we realize that we have a counterexample above, and since our logic is sound there can't be a proof of the formula; in particular it's this line (lemma B, line 8) of the proof that fundamentally can't be repaired.

Exercise 8

(Extra Credit 3 pts) Describe a board and an accompanying (simple) WFF where you know the formula is true, yet it's not possible to prove it via the WaterWorld domain axioms without mentioning propositions about the total-pirate-count totCount-is-1totCount-is-1, totCount-is-2totCount-is-2, …. (Indeed, you may have noticed that list of domain axioms makes only a paltry attempt at including domain axioms involving totCount-is-5totCount-is-5. There are quite a few omitted subcases as it is, even without considering other total-pirate-counts.) As always, a simpler example is better.

Suppose we don't patch up the WaterWorld domain axioms with numerous axioms for each of the total-pirate-count cases. What does your example above say about the soundness and the completeness of our WaterWorld domain axioms? (Recall that the pertinent problem when playing waterworld is not that you sometimes have to guess, but rather knowing whether a guess is the only option.)

Solution

Figure 2: A sample WaterWorld board
Figure 2 (z-safe.png)

Consider a board shown by the above figure, where there is only one pirate, and it's already been discovered at location Y. There are no domain axioms which let us deduce that Z-safeZ-safe. Observe that the proposition totCount-is-1totCount-is-1 is true for this board.

This means that our WaterWorld domain axioms are incomplete: there are true facts that cannot be deduced. In this case, it's possible to patch up our system by adding more domain axioms. (I'll leave it to you to think about how many rules you might need, for propositional logic.)

Relations and Interpretations

Exercise 9

(5 pts) Are each of the following formulas true for all interpretations ("valid")? (Remember that the relation-names are just names in the formula; don't assume the name has to have any bearing on their interpretation.)

  • For arbitrary aa,bb in the domain, atLeastAsWiseAsab ∨ atLeastAsWiseAsbaatLeastAsWiseAsab ∨ atLeastAsWiseAsba
  • For arbitrary aa in the domain, primea → odda → primeaprimea → odda → primea
  • For arbitrary aa,bb in the domain, betterThanab → ¬betterThanbabetterThanab → ¬betterThanba
For each, if it is true or false under all interpretations, prove that. (For these small examples, a truth table will probably be easier than using boolean algebra or an inference-system proof.) Otherwise, give and interpretation in which it is true, and one in which it is false.

Hint:

As always, look at trivial and small test cases first. Here, try domains with zero, one, or two elements, and small relations.

Solution

  • Not valid. The formula atLeastAsWiseAsab ∨ atLeastAsWiseAsbaatLeastAsWiseAsab ∨ atLeastAsWiseAsba is false when the domain is {appleapple} and the relation atLeastAsWiseAsatLeastAsWiseAs = {}. However it is true for the same domain but interpreting atLeastAsWiseAsatLeastAsWiseAs = {apple appleapple apple}.
  • Valid. Choose any interpretation you like, and we'll show that primea → odda → primeaprimea → odda → primea is true. Let PP be the truth value of primeaprimea under that interpretation, and similarly and let QQ be the truth value of oddaodda. Then (for that interpretation) the original formula has the truth value P → Q → PP → Q → P, which is a tautology back in propositional logic (truth table below), and so the original formula is true under all interpretations.
  • Not valid. The formula betterThanab → ¬betterThanbabetterThanab → ¬betterThanba is false when the domain is {appleapple} and betterThanbetterThan is interpreted as {apple appleapple apple}. This formula is true when the domain is the empty relation {}. (In general, it is true whenever betterThanbetterThan is interpeted as a antisymmetric relation which is nowhere reflexive.)
Table 8: Truth-table for propositional analog of a relational formula.
PPQQP → Q → PP → Q → P
falsefalsetrue
falsetruetrue
truefalsetrue
truetruetrue

Fortunately, none of these three formulas turned out to be unsatisfiable, else it sure would have been hard to give an example which where it did hold. (Taking the negation of a valid formula will give you an unsatisfiable one.)

Quantifiers

Exercise 10

(2 pts.) Rosen, Section 1.3, pg.40, #6. Give English translations of the following, where the domain is the students in your school, and N is the relation "has visited North Dakota".

  • ∃x. Nx∃x.Nx
  • ∀x. Nx∀x.Nx
  • ¬∃x. Nx ¬∃x. Nx
  • ∃x. ¬Nx∃x.¬Nx
  • ¬∀x. Nx¬∀x. Nx
  • ∀x. ¬Nx∀x.¬Nx

Solution

  • ∃x. Nx∃x.Nx : Some student in my school has visited North Dakota.
  • ∀x. Nx∀x.Nx: All students in my school have visited North Dakota.
  • ¬∃x. Nx ¬∃x. Nx : No student in my school has visited North Dakota.
  • ∃x. ¬Nx∃x.¬Nx : Some student in my school has not visited North Dakota.
  • ¬∀x. Nx¬∀x. Nx: Not all students in my school have visited North Dakota.
  • ∀x. ¬Nx∀x.¬Nx: All students in my school have not visited North Dakota.

Exercise 11

(3 pts.) Rosen, Section 1.3, pg.40, #10.

  • A student in your class has a cat, a dog, and a ferret.
  • All students in your class have a cat, a dog, or a ferret.
  • Some student in your class has a cat and a ferret but not a dog.
  • No student in your class has a cat, a dog, and a ferret.
  • For each of the three animals, there is a student in your class that has one.

Solution

  • ∃x. Cx ∧ Dx ∧ Fx∃x.Cx ∧ Dx ∧ Fx
  • ∀x. Cx ∨ Dx ∨ Fx∀x.Cx ∨ Dx ∨ Fx
  • ∃x. Cx ∧ Fx ∧ ¬Dx∃x.Cx ∧ Fx ∧ ¬Dx
  • ¬∃x. Cx ∧ Dx ∧ Fx ¬∃x. Cx ∧ Dx ∧ Fx
  • ∃x. Cx ∧ ∃y. Dy ∧ ∃z. Fz∃x. Cx ∧ ∃y. Dy ∧ ∃z. Fz

Exercise 12

(3 pts.) Rosen, Section 1.3, pg.41, #16.

  • ∃x. x2=2∃x.=x 2 2
  • ∃x. x2=-1∃x.=x 2 -1
  • ∀x. x2+2≥1∀x. x 2 2 1
  • ∀x. x2≠x∀x. x 2 x

Solution

  • true (as witnessed by each of {-sqrt(2), sqrt(2)}).
  • false.
  • true.
  • false (as witness by each of {0,1}).

Footnotes

  1. "transliterate" meaning a word-for-word substitution, while "translation" tries to preserve meanings/idioms. So, the German "Übung macht den Meister" transliterates to "Drill makes the master", but translates to "practice makes perfect".

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