160A Homework 3 Solutions - Due Wednesday Feb 8, 2012
1.7 #12 For each of the following conditions, give an example of an unsatisfiable set Γ of
formulas that meets the condition.
(a) Each member of Γ is - by itself - satisfiable.
Solution: Let γ1 = A1 and γ2 = ¬A1 and Γ = {γ1, γ2 }.
(b) For any two members γ1 , γ2 of Γ, the set {γ1 , γ2 } is satisfiable.
Solution: Let γ1 = A1 , γ2 = ¬A2 , γ3 = ¬A1 ∨ A2 and Γ = {γ1, γ2 , γ3 }. Then the
two element subsets of Γ are
– {A1 , ¬A2 }: satisfiable by v(A1 ) = T, v(A2 ) = F .
– {A1 , ¬A1 ∨ A2 }: satisfiable by v(A1 ) = T, v(A2 ) = T .
– {¬A2 , ¬A1 ∨ A2 }: satisfiable by v(A1 ) = F, v(A2 ) = F .
But, no truth assignment simultaneously satisfies all formulas in Γ.
1.7* A Prove that soundness is equivalent to the statement “every satisfiable set of formulas
is consistent.”
Solution: Recall that soundness is the statement: for all sets of wffs Σ and ϕ a wff,
Σ ϕ implies Σ ϕ.
Assume soundness. We will prove the contrapositive: if Σ is inconsistent then it is
not satisfiable. Suppose Σ is inconsistent and hence there is a wff ϕ such that Σ ϕ
and Σ ¬ϕ. Soundness implies that Σ ϕ and Σ ¬ϕ. Let v be a truth assignment
that satisfies every wff in Σ. By above, v̄(ϕ) = v̄(¬ϕ) = T . This is impossible. Hence,
no truth assignment satisfies every wff in Σ and Σ is unsatisfiable.
Conversely, suppose every satisfiable set of formulas is consistent and let Σ, ϕ be
as stipulated. We work towards a proof by contradiction: assume Σ ϕ but Σ ϕ.
Then by fact proved in class and on page 60 of Enderton, Σ ∪ {¬ϕ} is satisfiable.
Therefore (by hypothesis), Σ ∪ {¬ϕ} is consistent. Since Σ ϕ, Σ ∪ {¬ϕ} ϕ. Also,
since ¬ϕ ∈ Σ ∪ {¬ϕ}, Σ ∪ {¬ϕ} ¬ϕ. This contradicts the consistency of Σ ∪ {¬ϕ}.
1.7* B Prove that completeness is equivalent to the statement “every consistent set of formulas is satisfiable”.
Solution: Recall that completeness is the statement: for all sets of wffs Σ and ϕ a
wff, Σ ϕ implies Σ ϕ.
Assume completeness. Towards a contradiction, suppose Σ is a consistent set of
formulas that is not satisfiable. Since it is not satisfiable, it tautologically implies
anything. In particular, Σ A1 ∧ ¬A1 . By completeness assumption, this implies
that Σ A1 ∧ ¬A1 and let α0 , . . . , αn be a deduction that witnesses this. Note that
A1 ∧¬A1 → A1 and A1 ∧¬A1 → ¬A1 are both tautologies (can verify by truth-table).
Then
α0 , . . . , αn , A1 ∧ ¬A1 → A1 , A1 is a deduction that witnesses Σ A1 and
α0 , . . . , αn , A1 ∧ ¬A1 → ¬A1 , ¬A1 is a deduction that witnesses Σ ¬A1 . This contradicts the assumption that Σ is
consistent.
Conversely, suppose any consistent set is satisfiable. We will prove the contrapositive: that is, assume Σ ϕ and we will show that Σ ϕ. By Part (b) of fact proved
in lecture, Σ ∪ {¬ϕ} is consistent. The assumption then implies that Σ ∪ {¬ϕ} is
satisfiable. That is, there is a truth assignment which satisfies every wff in Σ and ¬ϕ.
This assignment satisfies every wff in Σ and does not satisfy ϕ. Therefore, Σ ϕ.
2.1 #2 Translate into good English the wff
∀x(Nx → Ix → ¬∀y(Ny → Iy → ¬x < y)).
We add in some parentheses dropped according to convention 4 from page 78:
∀x(Nx → Ix → ¬∀y(Ny → Iy → ¬x < y ) ).
Recall that ∃xα abbreviates ¬∀x(¬α) and therefore:
∀x(Nx → Ix → ∃y¬(Ny → Iy → ¬x < y ) ).
This is tautologically equivalent to
∀x(Nx → Ix → ∃y(Ny ∧ Iy ∧ x < y ) ).
Thus, all interesting numbers have some interesting number above them .
2.1 # 3 Translate into the first-order language specified: Neither a nor b is a member of every
set. (∀, for all sets; ∈, is a member of; a, a; b, b).
¬(∀x(a ∈ x) ∨ (∀x(b ∈ x)))