Allanic, Joshua G.
Math 51.1-C
Task 1. Assign each of the columns numbered 1 to 16 to a simplest formula (i.e., using as few characters as possible) using p, q,
the logical connectives ∧, ∨, →, ↔ , ¬, and/or the special statements . (e.g., Column 8 is ).​
​
p
q
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
F
F
T
F
T
F
T
F
T
F
T
F
T
F
T
F
T
F
F
T
T
T
F
F
T
T
F
F
T
T
F
F
T
T
F
F
T
F
T
T
T
T
F
F
F
F
T
T
T
T
F
F
F
F
T
T
T
T
T
T
T
T
T
T
F
F
F
F
F
F
F
F
Answers
1.​ p v ¬p
2.​ p ∨ q
3.​ p ∨ ¬q
4.​ p (p ∧ p ≡ p)
5.​ p → q
6.​ q (q ∧ q ≡ q)
7.​ p ↔ q
8.​ p ∧ q
9.​ ¬p ∨ ¬q
10.​ ¬(p ↔ q)
11.​ ¬q
12.​ p ∧ ¬q
13.​ ¬p
14.​ ¬p ∧ q
15.​ ¬(p ∨ q)
16.​ p ∧ ¬p
Task 2. A set of logical connectives is said to be functionally complete if it can be used to express all possible truth tables by
combining members of the set into a valid compound proposition.
(a)
We want to show that the set {∧, ∨, ¬} is functionally complete for two propositions.
1.​ Let p and q be the propositions.
2.​ Since there are 2 propositions, there are 4 possible combinations of truth values:
p
q
F
F
F
T
T
F
T
T
3.​ For each of these 4 rows, the output of a truth table can be either T or F. Thus, the number of truth tables is 24 or 16.
Task 1 allows listing all 16 possible truth tables for p and q.
p
q
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
F
F
T
F
T
F
T
F
T
F
T
F
T
F
T
F
T
F
F
T
T
T
F
F
T
T
F
F
T
T
F
F
T
T
F
F
T
F
T
T
T
T
F
F
F
F
T
T
T
T
F
F
F
F
T
T
T
T
T
T
T
T
T
T
F
F
F
F
F
F
F
F
4.​ Each of these truth tables was matched with a corresponding formula.
p
q
p v ¬p
p∨q
p ∨ ¬q
p
p→q
q
p↔q
p∧q
¬p ∨
¬q
¬(p ↔ q)
¬q
p ∧ ¬q
¬p
¬p ∧ q
¬(p ∨ q)
p∧
¬p
F
F
T
F
T
F
T
F
T
F
T
F
T
F
T
F
T
F
F
T
T
T
F
F
T
T
F
F
T
T
F
F
T
T
F
F
T
F
T
T
T
T
F
F
F
F
T
T
T
T
F
F
F
F
T
T
T
T
T
T
T
T
T
T
F
F
F
F
F
F
F
F
5.​ Most of the formulas used ∧, ∨, ¬ except for p → q, p ↔ q, and ¬(p ↔ q). However, any formula containing an
implication (→) or biconditional (↔) can be expressed using ∧, ∨, ¬. To show:
p → q ≣ ¬p ∨ q
by Switcheroo
p ↔ q ≣ (p → q) ∧ (q → p)
≣ (¬p ∨ q) ∧ (¬q ∨ p)
by Biconditional Definition
by Switcheroo
¬(p ↔ q) ≣ p ↔ ¬q
≣ (p → ¬q) ∧ (¬q → p)
≣ (¬p ∨ ¬q) ∧ (q ∨ p)
by Negation of biconditional
by Biconditional Definition
by Switcheroo
6.​ Thus, all 16 possible truth tables for p and q can be expressed using only ∧, ∨, and ¬.
7.​ Since every possible truth table for p and q can be expressed using only ∧, ∨, and ¬, the set {∧, ∨, ¬} is functionally
complete for two propositions.
However, the proof above can be only true when there are two propositions. We want to show that the set {∧, ∨, ¬} is
functionally complete regardless of the number of propositions.
1.​ Let n be the number of propositions.
2.​ Suppose n = 3.
3.​ Let s, r, t be the propositions.
4.​ We can uniquely construct a formula for any truth table of n propositions by identifying every individual row that
results in a True (T) output. This method is called Disjunctive Normal Form (DNF).
5.​ For example, consider a custom rule for s,r,t that outputs T in exactly three specific scenarios, as shown in the truth
table below:
s
r
t
Truth Value
F
F
F
F
F
F
T
F
F
T
F
T
F
T
T
F
T
F
F
F
T
F
T
T
T
T
F
F
T
T
T
T
6.​ For each row that outputs T, we isolate the inputs and construct a formula using only ∧ and ¬ to describe the exact
state:
a.​ Row 3: (¬s ∧ r ∧ ¬t)
b.​ Row 6: (s ∧ ¬r ∧ t)
c.​ Row 8: (s ∧ r ∧ t)
7.​ Since the overall function is T if Row 3, Row 6, or Row 8 occurs, we link their respective formulas together using ∨ to
build the complete compound proposition given in this formula:
(¬s ∧ r ∧ ¬t) ∨ (s ∧ ¬r ∧ t) ∨ (s ∧ r ∧ t)
8.​ This can be repeated for any possible truth table with n. Therefore, since every possible truth table can be mapped
into an equivalent DNF using only ∧, ∨, and ¬, the set {∧, ∨, ¬} is functionally complete regardless of the number of
propositions.
(b)
p
q
p★q
F
F
T
F
T
T
T
F
T
T
T
F
We want to show that the set { ★ } is functionally complete.
1.​ Let the operator ★ be defined by the given truth table, where p ★ q ≣ ¬(p ∧ q).
2.​ We can prove the set { ★ } is functionally complete by showing it has an equivalent operators of the already proven
set {∧, ∨, ¬}.
3.​ SECTION 1: We can replicate the negation (¬) by:
Proof:
1 p ★ q ≣ ¬(p ∧ q)
By Definition of ★
2 p ★ p ≣ ¬(p ∧ p)
Substituting p for q
3p∧p≣p
By Idempotent Law
4 p ★ p ≣ ¬p
By Substituting step 3 into step 2
4.​ SECTION 2: We can replicate the conjunction (∧) by:
p ★ q ≣ ¬(p ∧ q)
By Definition of ★
¬(p ★ q) ≣ ¬(¬(p ∧ q))
Negating both sides
¬(p ★ q) ≣ p ∧ q
By Double Negation Law
(p ★ q) ★ (p ★ q) ≣ ¬(p ★ q)
By Substituting the expression ( p ★ q) into the Negation
rule from Section 1
p ∧ q ≣ (p ★ q) ★ (p ★ q)
By Substitution
5.​ SECTION 3: We can replicate the disconjunction (∨) by:
p ∨ q ≣ ¬( ¬p ∧ ¬q)
By De Morgan's Law
¬(¬p ∧ q) ≣ ¬p ★ ¬q
By Definition of ★
p ∨ q ≣ ¬p ★ ¬q
By Substitution
p ★ p ≣ ¬p
From Negation rule in Section 1
q ★ q ≣ ¬q
From Negation rule in Section 1
p ∨ q ≣ (p ★ p) ★ (q ★ q)
By Substituting steps 4 and 5 into step 3
6.​ Since the set {∧, ∨, ¬} can all be expressed strictly using the ★ operator, the set { ★ } is functionally complete.