Propositional logic
Propositions: p: the sky is blue
p
q
~p
T
T
F
F
T
F
T
F
F
F
T
T
p and q
p or q
T
F
F
F
T
T
T
F
q: it is raining
p implies q p equiv q
T
F
T
T
T
F
F
T
A proposition can take one of two (True or False) truth values.
Logical operators: negation (not), and (conjunction), (inclusive) or (disjunction),
implication, equivalence, (exclusive) or, many others.
Logical equivalences
T and x = x
F or x = x
x and x = x
x or x = x
x and (y or z) = (x and y) or (x and z)
x or (y and z) = (x or y) and (x or z)
~~x=x
~ (x or y) = ~x and ~y
~(x and y) = ~x or ~y
x implies y = ~x or y
x equiv y = (x implies y) and (y implies x)
There are 16 propositional logic operators:
p
T
T
F
F
q
T
F
T
F
1
T
T
T
T
2
T
T
T
F
3
T
T
F
T
4
T
T
F
F
5
T
F
T
T
6
T
F
T
F
7
T
F
F
T
8 9
T F
F T
F T
F T
10
F
T
T
F
11
F
T
F
T
12
F
T
F
F
13
F
F
T
T
14
F
F
T
F
15
F
F
F
T
16
F
F
F
F
Question: what is the minimal set of logical operators required to
represent all of them?
Candidates: Disjunction, Conjunction, negation? Are these sufficient to represent all
operators? (yes, see later discussion of CNF and DNF). Do we need all of them? (no)
(x and y) = ~(~x or ~y) – that is, and can be expressed as or and ~, so or and ~ are sufficient minimal set
(x or y) = ~(~x and ~y) – that is, or can be expressed as and and ~, so and and ~ are sufficient minimal set
A smaller minimal set is nand : p nand q = ~(p and q)
and another minimal set is nor: p nor q = ~(p or q)
Show that each of the following is a minimal set of operators: {and, ~}, {or, ~}, {nand}, {nor}, {à, ~},
and that {~, ßà} is not a minimal set.
Logical equivalences can be verified using truth tables:
x equiv y = (x implies y) and (y implies x)
x y xày yàx
T
T
F
F
T
F
T
F
T
F
T
T
T
T
F
T
(xày) and (y à x)
x ßà y
T
F
F
T
T
F
F
T
Columns are identical
Using equivalences to simplify propositional formulas (e.g., reduce the
number literals or the number of terms, etc.)
p or (~p and q) = (p or ~p) and (p or q) = T and (p or q) = p or q
(p and q) or (~p and r) or (q and r)
= (p and q) or (~p and r) or ((q and r) and T)
= (p and q) or (~p and r) or ((q and r) and (p or ~p))
= (p and q) or (~p and r) or ((q and r) and p) or ((q and r) and ~p)
=…
= ((p and q) and (T or r)) or ((~p and r) and (T or q))
=…
= (p and q) or (~p and r)
A propositional formula or sentence is
satisfiable iff there is at least one assignment of truth values to
its propositions/literals that makes the sentence true.
((p and q) or (~p implies q)) and p
p
q
x
p and q
T
T
F
F
T
F
T
F
T
F
F
F
~p
y
~p implies q
x or y
F
F
T
T
T
T
T
F
T
T
T
F
(x or y) and p
T
T
F
F
At least on row is T
A propositional sentence is unsatisfiable (a contradiction) if no
assignment of truth values to the propositions make the
sentence true.
~ (((p or q) and ~q) implies p)) is unsatisfiable
p
q
p or q
~q
T
T
F
F
T
F
T
F
T
T
T
F
F
T
F
T
x
y
((p or q) and ~q) x à p
F
T
F
F
T
T
T
T
((p or q) and ~q) implies p is valid (a tautology)
~y
F
F
F
F
Inference rules
Modus ponens: x à y, x |- y
Whenever both x and xày are T, y is also T
x
T
T
F
F
y
T
F
T
F
xà y
T
F
T
T
x and xày
T
Modus ponens is sound
F
F
F
Modus tolens: x à y, ~y |- ~x
Modus tolens is sound
Proof: Suppose that we are given the following axioms:
p
r
pà q
(we can interpret this as (p and r and (p implies q))
p
Prove (q and r):
1. p, pàq |- q
2. r, q |- q and r
(modus ponens)
(and-introduction)
pàq
r
q
q and r
There are some things that are true that we may not be able to
prove with modus ponens (for example): modus ponens is not
complete.
Consider two axioms: xày, yàz . Prove xàz
x
y
z
xày
yàz
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
F
F
T
T
T
T
T
F
T
T
T
F
T
T
xàz
T
F
T
F
T
T
T
T
Can we prove this with modus ponens (along with and-introduction,
modus tolens, equivalences, etc.)? Modus ponens in not complete
(that is we cannot prove everything that logically follows from the
axioms with modus ponens alone.)
Resolution
x or y, ~y or z |- x or z
(x and/or z may by NULL)
If y=T then z; if y=F then x
Resolution subsumes modus tolens and modus ponens and some
inference rules. Resolution is sound.
Modus ponens can be reexpressed (remember xày = ~x or y):
xày, x |- y
è ~x or y, x |- y (resolve ~x and x)
Modus Tolens can be reexpressed:
xày, ~y |- ~x è ~x or y, ~y |- ~x (resolve y and ~y)
Proof by resolution:
Convert KB into clause form (and break into clauses)
e.g., convert ~(pàq) or (r à p)
a. Eliminate implication signs by using equivalent or expression
~(pàq) or (rà p) = ~(~p or q) or (~r or p)
b. push negations in and eliminate
~(~p or q) or (~r or p) = (~~p and ~q) or (~r or p)
= (p and ~q) or (~r or p)
c. Convert to CNF by using associate and distributive laws
(p and ~q) or (~r or p) = (p or ~r or p) and (~q or ~r or p)
= (p or ~r) and (~q or ~r or p)
d. Write as a list (conjunction) of clauses: (p or ~r), (~q or ~r or p)
A proof by resolution:
Given axioms:
p
(p and q) à r
(s or t) à q
axioms in clause form:
p
~p or ~q or r
~s or q
~t or q
t
t
Prove r using resolution only:
p
~p or ~q or r
~t or q
~q or r
resolvants
(theorems)
~t or r
r
t
Axioms used in proof
Axioms(KB): ~s or p
Prove: p and (q or r)
~u or s
u
~w or r
w
We cannot prove with resolution alone without refutation, since we
have no way of introducing q, though we can make use use of resolution
and other inference rules to show that that it must be true.
p must be true: ~u or s
u
~s or p r must be true: ~w or r
s
w
r
p
so (q or r) is true
by the inference
rule of or-introduction
so p and (q or r) is true by
inference rule of and-introduction
Axioms(KB): ~s or p ~u or s
u
~w or r
w
Prove: p and (q or r) (prove with resolution alone under refutation)
Negate (p and (q or r)) and convert to clause form:
~(p and (q or r)) = ~p or ~(q or r) = ~p or (~q and ~r)
= (~p or ~q) and (~p or ~r)
two clauses: (~p or ~q), (~p or ~r)
~w or r ~p or ~r
~s or p
~u or s
u
w
~w or ~p
Is there a simpler proof?
~w or ~s
~w or ~u
~w
{} Contradiction
Axioms = {~s or p, ~u or s, u, ~w or r, w}
NegatedSentence = {~p or ~q, ~p or ~r }
Search for a resolution-refutation proof
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
Axioms = {~s or p, ~u or s, u, ~w or r, w}
NegatedSentence = {~p or ~q, ~p or ~r }
Search for a resolution-refutation proof
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……………..
Axioms = {~s or p, ~u or s, u, ~w or r, w}
NegatedSentence = {~p or ~q, ~p or ~r }
Search for a resolution-refutation proof
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……………..
(p or ~s or ~u) ……. (p or ~u) …………………………………………… ~w ……………..
Axioms = {~s or p, ~u or s, u, ~w or r, w}
NegatedSentence = {~p or ~q, ~p or ~r }
Search for a resolution-refutation proof
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……………..
(p or ~s or ~u) ……. (p or ~u) …………………………………………… ~w ……………..
NULL (Contradiction)
Breadth-first search: start with 0-level resolvants, I = -1
do I=I+1
Initialize I+1 level resolvants
for each I-level resolvant, R
if R is contradiction then exit (proposition true)
for each other J-level resolvant (J <= I), S
add resolve(R,S) to I+1 level resolvants
until no change in I and I+1 level resolvants
exit (proposition false)
Depth-first search(0-level resolvants)
for each resolvant, R
P = DFS(R, 0-level resolvants)
if P indicates success (contradiction found) then exit success
exit failure
DFS(R, resolvants)
if R a contradiction then exit/return (proposition true)
for each resolvant, S
x=resolve(R,S)
if x NOT IN resolvants
P = DFS(x, resolvants+x)
if P indicates success (contradiction found) then exit success
exit failure (no contradiction found)
Illustration of depth-first search for proof
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……………..
(p or ~s or ~u) ……. (p or ~u) …………………………………………… ~w ……………..
NULL
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……………..
(p or ~s or ~u) ……. (p or ~u) …………………………………………… ~w ……………..
NULL
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……………..
(p or ~s or ~u) ……. (p or ~u) …………………………………………… ~w ……………..
NULL
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……………..
(p or ~s or ~u) ……. (p or ~u) (p or ~s)……………………………….. ~w ……………..
NULL
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……………..
(p or ~s or ~u) ……. (p or ~u) (p or ~s)………(p or ~u or ~s)………… ~w ……………..
NULL
After much search:
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……………..
(p or ~s or ~u) ……. (p or ~u) …………………………………………… ~w ……………..
NULL
After more search:
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……~u or ~r………..
(p or ~s or ~u) ……. (p or ~u) …………………………………………… ~w ……………..
NULL
After more search:
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……~u or ~r………..
(p or ~s or ~u) ……. (p or ~u) …………………………………………… ~w ………~r……..
NULL
After more search:
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……~u or ~r………..
(p or ~s or ~u) ……. (p or ~u) …………………………………………… ~w ………~r……..
NULL
~w
After more search:
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……~u or ~r………..
(p or ~s or ~u) ……. (p or ~u) …………………………………………… ~w ………~r……..
NULL
~w
NULL
You might be tempted to think from the last example, that in a DFS we need only
resolve the current clause (R is pseudocode earlier) with the 0-level clauses
(axioms + negated sentence). If you restricted yourself to this, then your algorithm
would certainly not be complete.
Consider the following axioms ~p or q, p or ~q, ~p or ~q
Prove: ~p and ~q Negation: ~(~p and ~q) equivalent to (p or q)
Attempt a DFS proof in which the current clause is always paired with a level-0
clause
p or q
~p or q
p or ~q ~p or ~q
q
p
~q
~p Start repeating and conclude no proof after much search
You might be tempted to think from the last example, that in a DFS we need only
resolve the current clause (R is pseudocode earlier) with the 0-level clauses
(axioms + negated sentence). If you restricted yourself to this, then your algorithm
would certainly not be complete.
Consider the following axioms ~p or q, p or ~q, ~p or ~q
Prove: ~p and ~q Negation: ~(~p and ~q) equivalent to (p or q)
Attempt a DFS proof in which the current clause is always paired with a level-0
clause
p or q
~p or q
p or ~q ~p or ~q
q
p or q
~p or q
p or ~q ~p or ~q
q
p
p
~q
~q
~p
Start repeating and conclude no proof after much search
NULL
These in resolvants
Set too
Set of support strategy: at least one resolvant is a clause of the original negated
sentence or a descendent of such a clause. The DFS-generated proofs of the
previous 3 pages satisfy the set-of-support constraint. The proof below does
NOT satisfy the set of support strategy because s is a result of resolving
two statements that are NOT from the set of clauses following from the negated
statement.
{~s or p, ~u or s, u, ~w or r, w, ~p or ~q, ~p or ~r } 0-level resolvants
(p or ~u) (~s or p or u) (~s or p or ~w or r) (~s or p or w) (~s or ~q) (~s or ~r) s … (~p or ~q or ~r)
(p or ~s or ~u) (p or ~u or s) p … (p or ~s) (p or ~u or ~s or ~w or r) …. ~r ……………..
(p or ~s or ~u) ……. (p or ~u) …………………………………………… ~w ……………..
NULL
Depth-first search(0-level resolvants)
for each resolvant, R
P = DFS(R, 0-level resolvants)
if P indicates success (contradiction found) then exit success
exit failure
Distinguish axioms and clauses of negated sentence and
Axioms and set-of-support
DFS(R, resolvants)
if R a contradiction then exit/return (proposition true)
for each resolvant, S
x=resolve(R,S)
Add x to set-of-support for recursive calls to DFS, but this addition needs to be retracted
after the recursive call terminates
if x NOT IN resolvants
P = DFS(x, resolvants+x)
if P indicates success (contradiction found) then exit success
exit failure (no contradiction found)
Return proof steps, not simply success or failure