Logic HW1
Elias Timothy
August 2026
Problem 1
Proposition for leftmost tree is (¬¬¬ ⊥). Proposition for middle tree is ((p0 =⇒
⊥) =⇒ ((p0 ⇐⇒ p1 ) ∧ p5 )). Proposition for rightmost tree is ¬(¬p1 =⇒
¬p1 ).
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Problem 2
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Problem 3
a. Let Γ = {φ}. Let v be an arbitrary valuation, if ∥φ∥v = 0, then every
element of Γ evaluates to 0 and the statement is true. If ∥φ∥v = 1(which is
every element of Γ), then ∥φ∥v = 1, satisfying the definition that Γ |= φ.
b. Assume φ |= ψ, and ψ |= σ.Let v be an arbitrary valuation, if ∥φ∥v = 1,
then ∥ψ∥v = 1. If ∥ψ∥v = 1 ,then ∥σ∥v = 1. Thus, if ∥φ∥v = 1 then ∥σ∥v = 1.
If ∥φ∥v = 0, the condition vacuously holds.
c. Assume |= φ =⇒ ψ. Then for any valuation v, if ∥φ∥v = 1 , then
∥ψ∥v = 1, thus φ |= ψ. Now assume φ |= ψ. Consider an arbitrary valuation
v, if ∥φ∥v = 0, then ∥φ =⇒ ψ∥v = 1 . If ∥φ∥v = 1, then by our hypothesis,
∥ψ∥v = 1, thus ∥φ =⇒ ψ∥v = 1. It follows that |= φ =⇒ ψ holds, completing
the proof.
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Problem 4
Consider atom p1 . Consider 1-ary connective ϕ with truth function f : {0, 1} →
{0, 1}, f (0) = 1, f (1) = 1. Note that the possible 1-ary connectives ¬ can form
are p1 and ¬p1 , because p1 , ¬p1 , ¬¬p1 , ... is a sequence with period 2 due to
¬¬p1 = p1 .Thus if {¬} were to be a functionally complete set of connectives,
then either p1 has the same truth function as ϕ, or ¬p1 has the same truth
function as ϕ. However, it can be verified in either case that the truth tables
do not match, hence {¬} is not a functionally complete set of connectives.
To show { =⇒ , ∨} does not form a logically complete set of connectives,
consider 1-ary connective φ with truth function f : {0, 1} → {0, 1}, f (0) =
0, f (1) = 0. Consider an arbitrary proposition p1 . Then for any valuation v,
if ∥p1 ∥v = 1, then both ∥p1 ∨ p1 ∥v = 1, and ∥p1 =⇒ p1 ∥v = 1. Under
this valuation, it can be shown that for any proposition τ built form p1 and
{ =⇒ , ∨}, we will have ∥τ ∥v = 1. However, f requires f (∥p1 ∥v ) = 0, thus τ is
not equivalent to φ. Since τ was arbitrary , we can conclude { =⇒ , ∨} does
not form a logically complete set of connectives.
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Problem 5
Recall that , |= ¬φ ⇐⇒ φ | φ. Additionally, for every value of p, q, ¬p | ¬q
is false only when both p and q are false, otherwise it is true. Similarly, p ∨ q.
is false only when both p and q are false, otherwise it is true. Thus, |= ¬p |
¬q ⇐⇒ p ∨ q. Since {¬, ∨} form a functionally complete set of connectives and
both connectives can be formed in terms of |, it follows that {|} is a functionally
complete set of connectives.
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Problem 6
Since ∧ may be expressed in terms of ∨ and ¬(by De Morgans laws), we will
use ∧ freely. It suffices to show |= $ ⇐⇒ [(p1 ∧ p2 ) ∨ (p2 ∧ p3 ) ∨ (p1 ∧ p3 ])]. If
$ is true, then at least two of p1 , p2 , p3 are true. WLOG let p1 , p2 both be true,
then p1 ∧ p2 = ¬ ⊥. Since ∥(¬ ⊥) ∨ q2 ∨ q3 ∥ = 1 for any propositions q2 and q3 ,
[(p1 ∧ p2 ) ∨ (p2 ∧ p3 ) ∨ (p1 ∧ p3 ])] must be true. If $ is false, then at most 1 of
p1 , p2 , p3 is true, and thus every pi ∧ pj is false. Since ∥ ⊥ ∨ ⊥ ∨ ⊥ ∥ = 0, it
follows that [(p1 ∧ p2 ) ∨ (p2 ∧ p3 ) ∨ (p1 ∧ p3 ])] is false, completing the proof.
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Problem 7
The disjunctive normal form is (¬φ ∧ ψ) ∨ (φ ∧ ¬ψ). The normal conjunctive
form is (φ ∨ ψ) ∧ (¬φ ∨ ¬ψ).
Second formula has conjunctive normal form ¬φ ∨ ψ and disjunctive normal
form (¬φ ∧ ¬ψ) ∨ (φ ∧ ψ) ∨ (¬φ ∧ ψ)
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