M (Hadley, Labrador) → ¬M(Chinook, Husky)
(M (Hadley, Labrador) ⊕ M (Hadley, Poodle)) ∧ [(M (Chinook, Husky) ⊕ M (Chinook, Poodle)) → ¬M (Hadley, Poodle)]
∀d:∃b:M (d, b)
"If Chinook is a Husky, then Hadley is a Labrador, and if Hadley is a Labrador, then Chinook is a Husky." Equivalently, "Chinook is a Husky if and only if Hadley is a Labrador."
"If two different dogs are members of a breed, then that breed is Labrador."
"For any breed, there do not exist a dog and a different breed such that the dog is a member of both breeds."
A function must be total and well-defined. Statement (c) says exactly that M is total. Statement (f) is almost exactly the statement that M is well-defined. If we use the negation-of-quantifier rule twice, and use commutativity of ∀ quantifiers, we can transform (f) into the equivalent statement ∀d:∀b1:∀b2:[M (d, b1) ∧ M (d, b2)] → (b1 = b2). This latter statement is exactly the definition of "M is well-defined".
M is not one-to-one because we will show in Question 4 that Chinook is neither a Husky nor a Poodle, and so Chinook is a Labrador by (c). We are given that Ebony is also a Labrador, and a one-to-one function could not have two different dogs with the same breed.
M is not onto, because none of the given dogs is a Husky and an onto function would have to have at least one dog of each breed. We will show below that Hadley is a Poodle.
Call the two booleans in question CH and HL. Statement (a) says
HL → ¬ CH, and statement (d) says CH ⊕ HL. We make a
two-variable truth table:
The truth table shows that the premises can only be satisfied if
HL and CH are both false.
(HL --> not CH) and (CH --> HL) and (HL --> CH)
0 1 1 0 1 0 1 0 |1| 0 1 0
0 1 0 1 1 1 0 0 |0| 0 1 1
1 1 1 0 1 0 1 1 |0| 1 0 0
1 0 0 1 0 1 1 1 |0| 1 1 1
We know ¬HL and ¬CH from statements (a) and (d), and these give
us no further information about the other two propositions, CP and HP.
Statement (b) says (HL ⊕ HP) ∧ ((CH ⊕ CP) → ¬HP).
Statement (e) affects our situation because it says that no two different
dogs can be Poodles, so we get ¬(CP ∧ HP).
By left separation on (b) we get HL ⊕ HP, and since we know that
HL is false, HP must be true. By right separation from (b) we get an
implication, and the contrapositive of this implication is
HP → (CH ↔ CP). (Here we use the rule (p ↔ q) ↔ ¬
(p ⊕ q).) Since HP is true, by modus ponens we get CH ↔ CP.
Since CP cannot be true by (e) (since HP is already true), CH and CP
cannot be both true, so
they are both false. Thus
we have ¬HL ∧ HP ∧ ¬CH ∧ ¬CP.
We can check that all our premises are satisfied by this truth
assignment: (a) is true vacuously, the left half of (b) is satisfied by
HP, the right half of (b) is true vacuously, (d) is true since CH and HL
are both false, and (e) is true because we have not assigned two Poodles.
Statement (f) says that no dog has two different breeds. Our assignment to the four booleans could have violated it by setting both HL and HP true, or both CH and CP true, but it didn't. The entire set of premises is satisfied if Chinook and Ebony are Labradors and Hadley is a Poodle.
A partial order is reflexive, antisymmetric, and transitive.
Pre is reflexive because ∀w:Pre(w,w) is true. To prove this,
let w be an arbitrary string. Pre(w,w) means ∃x:wx=w. This is
true by the Rule of Existence because we may take x to be λ. Since
w was arbitrary, we have proved the reflexive property.
The antisymmetry property is ∀u:∀v:(Pre(u,v)∧Pre(v,u))
→ (u=v). To prove this, let u and v be arbitrary strings and assume
Pre(u,v) and Pre(v,u). So we know ∃x:ux=v and ∃y:vy=u.
Consider the lengths |u| and |v| of u and v respectively.
Since |u| + |x| = |v|, we know |u| ≤ |v|,
and since |v| + |y| = |u|, we know |v| ≤ |u|. By properties of the
naturals, |u| must equal |v|. So x and y each have length 0, and so each
must be the empty string and we have that u = v. We have proved the
implication by direct proof. Since u and v were arbitrary, we have proved
the symmetric property.
The transitive property is ∀u:∀v:∀w:[Pre(u,v) ∧
Pre(v,w)] → Pre(u,w). So let u, v, and w be arbitrary strings and assume
Pre(u,v) and Pre(v,w). By the definition of Pre we have that
∃x:ux=v and ∃y:vy=w. Let x and y be specific strings satisfying
ux = v and vy = w, by the Rule of Instantiation. Substituting ux for v in
the equation vy = w, we find that uxy = w. Letting z by xy, we find that
∃z:uz=w which is the definition of Pre(u,w). We have proved the
implication, and since u, v, and w were all arbitary, we have proved the
transitive property.
Let a, b, c, and d be arbitrary naturals and assume D(a,b) and
c ≡ d (mod b). By the definition of division we know ∃e:ae=b,
so by Instantiation let e be a natural such that ae = b. By the given
definition of congruence we know that ∃q:(c=qb+d)∨(d=qb+c).
Let q be a natural such that c = qb+d or d = qb+c.
We prove c ≡ d (mod a) by cases. First assume that c = qb+d.
Since b = ae, we have that c = qae+d = a(qe)+d. By joining, we have
(c=a(qe)+d) ∨ (d=q(ae)+c). Letting z be qe, we have
∃z:(c=az+d) ∨
(d=az+c), which is the definition of c ≡ d (mod a).
For the other case, we assume d = qb + c. Again since b = ae we have
d = a(qe) + c, and by joining we get (d = a(qe) + c) ∨ (c = a(qe) + c).
Letting z = qe, we get ∃z:(d = az+c) ∨ (c = az+d) by the Rule of
Existence, and this is the definition of c ≡ d (mod a). We have
completed the proof by cases and thus proved the implication. Since a, b,
c, and d were arbitrary we have proved the desired statement.
Last modified 1 October 2004