PHILOSOPHY 2340
Symbolic Logic
THIRD EXAMINATION
Spring, 2001
Part I: Translations
Please translate the following sentences into FOL, using the Tarski's World abbreviations unless otherwise
indicated. (Note: for these exercises, interpret "some" to mean "at least one.") (5 points each.)
1. Some cube is in back of every tetrahedron.
∃x (Cube(x) ∧ ∀y (Tet(y) → BackOf(x,y)))
2. Some tetrahedron is to the left of everything else.
∃x (Tet(x) ∧ ∀y (y # x → LeftOf(x, y)))
3. Every farmer who owns a donkey beats it. (Use the predicates Farmer(x), Owns(x,y),
Donkey(x), Beats(x,y) – not necessarily with the same variables, of course!)
∀x ∀y ((Farmer(x) ∧ Donkey(y) ∧ Owns(x, y)) → Beats(x,y))
or
∀x (Farmer(x) → ∀y ((Donkey(y) ∧ Owns(x, y)) → Beats(x,y)))
4. Every tetrahedron with a cube to its right is larger than it (the cube, that is).
∀x ∀y ( (Tet(x) ∧ Cube(y) ∧ RightOf(y,x)) → Larger(x,y))
5. All cubes that are larger than everything else are either medium or to the left of a
tetrahedron. (to the left of some tetrahedron or other, that is)
∀x ( (Cube(x) ∧ Ay (y ≠ x → Larger(x,y)) ) → (Medium(x) ∨ ∃z (Tet(z) ∧
LeftOf(x,z)))
6. No tetrahedron is larger than every cube.
∀x (Tet(x) → ¬∀y (Cube(y) → Larger(x,y)) )
7. No cube with nothing in back of it is smaller than another cube.
∀x ( (Cube(x) ∧ Ay ¬BackOf(y,x)) → ¬∃z (Cube(z) ∧ z ≠ x ∧ Smaller(x,z)))
8. At least two cubes are smaller than a large tetrahedron. [NOTE: this one is ambiguous;
please give translations for both readings.]
∃x ∃y (x ≠ y ∧ Cube(x) ∧ Cube(y) ∧ ∃z (Large(z) ∧ Tet(z) ∧ Smaller(x,z) ∧
Smaller(y,z)))
∃x (Cube(x) ∧ ∃y (Large(y) ∧ Tet(y) ∧ Smaller(x,y)) ∧ ∃z (z ≠ x ∧ Cube(z) ∧ ∃w
(Large(w) ∧ Tet(w) ∧ Smaller(z,w)))
Part II: Axiomatization
Consider the location predicates FrontOf(x,y), BackOf(x,y), and SameRow(x,y). Write axioms to capture
the logical relationships between these predicates in Tarski’s World. Try to be complete – that is, try to
capture all the logical relations between these predicates in Tarski’s World. (There aren’t all that many
relations to worry about – depending on how you approach it, you could give a reasonable answer that
consisted of only three or four axioms. On the other hand, if you really wanted to, you could stretch it out
to about nine.) (10 points)
∀x ∀y (FrontOf(x,y) ∨ BackOf(x,y) ∨ SameRow(x,y))
∀x ∀y (FrontOf(x,y) → ¬(BackOf(x,y) ∨ SameRow(x,y))
∀x ∀y (BackOf(x,y) → ¬(FrontOf(x,y) ∨ SameRow(x,y))
∀x ∀y (SameRow(x,y) → ¬(BackOf(x,y) ∨ FrontOf(x,y))
(If the last three axioms were biconditionals, the first would be redundant)
Part III: Proofs
For the problems in Part III, you may use any of the following principles in addition to the introduction and
elimination rules: the DeMorgan's equivalences (for propositional logic – not the DeMorgan’s
equivalences for quantifiers!); Disjunctive Syllogism (from P  Q and P, infer Q); Excluded Middle
(write any sentence of the form P v P at any point in a proof); Conditional Exchange (from P  Q infer
P  Q and vice versa). As justification just write “taut con:” followed by “DM,” “DS,” “EM,” or “CE”
depending on which principle you use, followed by the appropriate line numbers (none needed for EM, of
course). (Don’t think that because I have listed these principles, you ought to try to use them, though! I
didn’t use any of them in doing the proofs when I arrived at the numbers of steps given below; I don’t
know for sure whether they could shorten any of the proofs or not.)
1. Prove that LeftOf(a,b)  xSmall(x) from the following premises. (Took me 9
steps.) 10 points.
1. xCube(x)  Small(b)
2. x (LeftOf(x, b)  Cube(x))
2. Here’s a proof you may find of interest. We mentioned some time ago that a universal
affirmative sentence is FO-equivalent to the negation of a particular negative sentence,
i.e. that x(S(x)  P(x)) is FO-equivalent to x(S(x)  P(x)). Since FO logic is
complete, we know that these two sentences must also be interderivable, that is, that
taking either one of them as our premise we can prove the other. Show this by
constructing two proofs. (Took me 10 steps for the first and 9 for the second.) (25 points.)
[first proof]
[second proof]
3. Prove xSmall(x) from the following premises. (Took me 17 steps.) 15 points.