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PHL 245: Practice Test: Second Test Focus on Units 4-6 UPCOMING TEST WILL BE (roughly): 40%: derivations 35%: symbolization A1: a advertises. D1: a has diamonds on it. G1: a is a restaurant M2: a is getting married to b.
B1: E1: H1: N2:
a0: Tiffany’s d 1: the fiancé of a
b0: Brian e1: the sister of a
1.
25%: concepts and truth-tables
is a business a is an engagement ring. a stays in business. a is more expensive than b. a
C1: F1: L2: O3:
a is
a store. a is a person. a is displayed at b. a buys b from c (c sells b to a).
c0: Carol b2: the best man at the wedding of a and b.
Use the above symbolization scheme to symbolize the following sentences: (There are more here for extra ex tra practice than will be on the second test.) (a) Although stores and restaurants are businesses, not all businesses advertise. ∀x(Cx ∨ Gx → Bx) ∧ ~∀ ~∀x(Bx → Ax)
OR ∀x(Cx → Bx) ∧ ∀x(Gx → Bx) ∧ ∃x(Bx ∧ ~Ax)
(b) Assuming that people don’t buy things from businesses that d on’t advertise, restaurants stay in business only if they do. ∀x(Bx ∧ ~Ax → ~∃ ~∃y∃z(Fy ∧ O(yzx))) → ∀x(Gx → (Hx → Ax)) or ~∃x (Bx ∧ ~Ax ∧ ∃y∃z(Fy ∧ O(yzx))) → ∀x(Gx ∧ Hx → Ax)
(c) In order for Carol’s fiancé to get married to Carol, it is necessary that he buys her an engagement ring from Tiffany’s. M(d(c)c) → ∃x(Ex ∧ O(d(c)xa)) or M(cd(c)) for the first part. (d) Stores that display engagement rings sell things with diamonds on them to people. ∀x(Cx ∧ ∃y(Ey ∧ L(yx)) → ∃z(Dz ∧ ∃w(Fw ∧ O(wzx)))) or ∀x(Cx → (∃ (∃y(Ey ∧ L(yx)) → ∃z(Dz ∧ ∃w(Fw ∧ O(wzx))))) or ∀x(Cx → ∀y(Ey ∧ L(yx) → ∃z(Dz ∧ ∃w(Fw ∧ O(wzx))))) Watch the effect of brackets on the quantifier for y (See (See confinement in unit 5 part 1: 5.10) (e) Everyone buys things from stores, but no store sells things to everyone. ∀x(Fx → ∃y∃z(Cz ∧ O(xyz))) ∧ ~∃ ~∃x(Cx ∧ ∀y(Fy → ∃zO(yzx)) OR second conjunct: ∀x(Cx → ∃y (Fy ∧ ~∃ ~∃z O(yzx)))
1
A1: a advertises. D1: a has diamonds on it. G1: a is a restaurant M2: a is getting married to b.
B1: E1: H1: N2:
a0: Tiffany’s d 1: the fiancé of a
b0: Brian e1: the sister of a
is a business a is an engagement ring. a stays in business. a is more expensive than b. a
C1: F1: L2: O3:
a is
a store. a is a person. a is displayed at b. a buys b from c (c sells b to a).
c0: Carol b2: the best man at the wedding of a and b.
(f) The only store that Brian buys an engagement ring from is Tiffany’s. ∀x(Cx ∧ ∃y(Ey ∧ O(byx)) ↔ x=a) OR ∃x(∀y(Cy ∧ ∃z(Ez ∧ O(bzy)) ↔ x=y) ∧ x=a)
(g) If anybody buys an engagement ring from a store then he/she is getting married to somebody. ∀x(Fx ∧ ∃y(Ey ∧ ∃z(Cz ∧ O(xyz))) → ∃w(Fw ∧ M(xw))
(h) The best man at Carol and Brian’s wedding is Carol’s sister’s fiancé. a(bc)=d(e(c))
(i) Give an idiomatic English translation of: ∃x(Cx ∧ ∀y(Cy ∧ x ≠ y → N(xy)) ∧ ∀z(L(zx) → Dz)). Everything displayed at the most expensive store has diamonds on it.
(j) Disambiguate this ambiguous sentence by providing two symbolizations. For each, provide an English sentence that makes the meaning clear. Everybody buys something from a store. The two natural interpretations are: ∀x(Fx → ∃y∃z(Cz ∧ O(xyz)))
Everybody buys something from some store or another. (diff. things, diff. stores)
∃z(Cz ∧ ∀x(Fx → ∃yO(xyz)))
There is one store that everybody buys something from. (diff. things, same store)
Two other (less plausible) interpretations: ∃y∀x(Fx → ∃z(Cz ∧ O(xyz)))
Everybody buys the same one thing from some store or another. (same thing, diff. stores)
∃y∃z(Cz ∧ ∀x(Fx → O(xyz)))
Everybody buys the same thing from the same store. (same thing, same store)
2
3. Use a full truth table to determine whether the following is a tautology, a contradiction or a contingent sentence. State which it is and briefly explain how you know. P → (Q ∨ R) ∧ ~(P ↔Q) ↓ P
Q
R
P
T T T T F F F F
T T F F T T F F
T F T F T F T F
T T T T F F F F
(Q
F F T F T T T T
T T F F T T F F
R)
T T T F T T T F
T F T F T F T F
F F T F T T F F
~
(P
F F T T T T F F
T T T T F F F F
Q)
T T F F F F T T
T T F F T T F F
← false on this TVA ← true on this TVA
This is a contingent sentence since some lines below the main connective are true and some false.
4. Use a shortened truth-table of one line to show that the following argument is INVALID. P → Q ∨ R. P
Q
R
S
T
T
F
T
↓
~(Q ↔ ~S ∧ P).
∴ P → R.
↓
P
Q
R
T T
T
T F
.
↓
~
(Q
T
T
F
~
S
P)
P
R
F
T F
T
T F
F
The premises are both true, but the conclusion is false. Therefore, the argument is INVALID.
3
5. Show that the following arguments are valid:
a)
∀y(Fy → ∃z(Jz ∧ Gz)).
∃x(Jx ∨ Bx) → ∀x∀yH(xy).
∴∀x(Fx → ∃y(Gy ∧ H(xy)))
Tests basic skills with UD, EG, EI and UI.
4
b) ∃x∀y(H(xyy) →∀z~L(xz)). ∀x( ∃z(Gz ∧ ~Mz) → K(xx)).
A bit tougher… You will need to see that you have to derive the antecedent of premise 2 (or the negation of the consequent.) Then it is all about matching. Always use a new term for EI (ASAP… use a brand new term.) Use UI to match (always match a free term… often the arbitrary term from EI or from setting up a UD.) Use EG to match a bound variable (often in show lines, consequents of show lines and antecedents of lines that you haven’t used.)
