Introduction & propositional calculus
First-order logic
Miscellaneous
Introduction & propositional calculus
First-order logic
Miscellaneous
Lecture II: tableaux, satisfiability
Lecture III: resolution
Propositional calculus: tableaux
Logic: Compendium
http://www.ida.liu.se/∼TDDD72/
Andrzej Szalas
IDA, University of Linköping
May 23, 2016
Andrzej Szalas
Introduction & propositional calculus
First-order logic
Miscellaneous
Semantic tableau for a formula A
A semantic tableau T is a tree with each node labeled with
a set of formulas, where T represents A in such a way that
A is equivalent to the disjunction of formulas appearing in all
leaves, assuming that sets of formulas labeling leaves are
interpreted as conjunctions of their members.
Slide 1 of 43
Andrzej Szalas
Lecture II: tableaux, satisfiability
Lecture III: resolution
α-formulas
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Slide 2 of 43
Lecture II: tableaux, satisfiability
Lecture III: resolution
β-formulas
Rule
(¬¬)
(∧)
(¬∨)
(¬ →)
(↔)
α
¬¬A1
A1 ∧ A2
¬(A1 ∨ A2 )
¬(A1 → A2 )
A1 ↔ A 2
α1
A1
A1
¬A1
A1
A1 → A2
Factorization
Rule (fctr): remove redundant duplicates.
α2
A2
¬A2
¬A2
A2 → A1
Rule
β
(¬∧) ¬(B1 ∧ B2 )
(∨)
B1 ∨ B2
(→)
B1 → B2
(¬ ↔) ¬(B1 ↔ B2 )
β1
¬B1
B1
¬B1
¬(B1 → B2 )
E.g. p, q, p, r , r is simplified to p, q, r .
Andrzej Szalas
Slide 3 of 43
Andrzej Szalas
Slide 4 of 43
β2
¬B2
B2
B2
¬(B2 → B1 )
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Literals and complementary formulas
A literal is an atom (propositional variable) or the negation
of an atom. Atoms are called positive literals and their
negations are called negative literals.
For any formula A, {A, ¬A} is a complementary pair
of formulas.
A is the complement of ¬A and ¬A is the complement of A.
Introduction & propositional calculus
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Closed and open leaves
A leaf is called closed if it contains a complementary
pair of formulas.
If a leaf consists of literals only and contains no
complementary pair of literals then we call it open.
A tableau is completed if all its leaves are open or closed.
Remark
Observe that the conjunction of a complementary pair of formulas
is equivalent to false.
Andrzej Szalas
Introduction & propositional calculus
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Slide 5 of 43
Lecture II: tableaux, satisfiability
Lecture III: resolution
Construction of a semantic tableau for a formula C
Initially, T consists of a single node labeled with {C }.
If T is completed then no further construction is possible.
Otherwise chose a leaf, say l, labeled with S containing
a non-literal and chose from S a formula D which is not
a literal and:
if D is! an α-formula
then create a successor of l and label it
with S − {D} ∪ {α1 , α2 }
Andrzej Szalas
Introduction & propositional calculus
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Slide 6 of 43
Lecture II: tableaux, satisfiability
Lecture III: resolution
Proving with semantic tableaux
To prove that a formula A is a tautology we construct
a completed tableau for its negation ¬A.
If the tableau is closed then A is a tautology
(its negation is not satisfiable).
Otherwise A is not a tautology.
if D is a β-formula then
! create two
new successors of l, the
first one labeled
with
S
−
{D}
∪ {β1 } and the second one
!
labeled with S − {D} ∪ {β2 }.
Andrzej Szalas
Slide 7 of 43
Andrzej Szalas
Slide 8 of 43
Introduction & propositional calculus
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NNF: Negation Normal Form
A literal of the form A is called positive
and of the form ¬A is called negative.
We say that formula A is in negation normal form,
abbreviated by Nnf, iff it contains no other connectives than
∧, ∨, ¬, and the negation sign ¬ appears in literals only.
Introduction & propositional calculus
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Any propositional formula can be equivalently transformed into the
Nnf by replacing subformulas according to the table below, until
Nnf is obtained.
Rule Subformula
Replaced by
1
A↔B
(¬A ∨ B) ∧ (A ∨ ¬B)
2
A→B
¬A ∨ B
3
¬¬A
A
4
¬(A ∨ B)
¬A ∧ ¬B
5
¬(A ∧ B)
¬A ∨ ¬B
Slide 9 of 43
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Lecture II: tableaux, satisfiability
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CNF: Conjunctive Normal Form
A clause is any formula of the form A1 ∨ A2 ∨ . . . ∨ Ak , where
k ≥ 0 and A1 , A2 , . . . , Ak are literals.
The empty clause (when k = 0), denoted by ∅, is equivalent
to F.
A Horn clause is a clause in which at most one literal is
positive.
Formula A is in conjunctive normal form, abbreviated by Cnf,
if it is a conjunction of clauses. It is in clausal form if it is
a set of clauses (considered to be an implicit conjunction of
clauses).
Andrzej Szalas
Lecture II: tableaux, satisfiability
Lecture III: resolution
Transforming formulas into NNF
Recall that a literal is a formula of the form A or ¬A,
where A is an atomic formula.
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Introduction & propositional calculus
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Lecture II: tableaux, satisfiability
Lecture III: resolution
Slide 11 of 43
Slide 10 of 43
Lecture II: tableaux, satisfiability
Lecture III: resolution
Transforming formulas into CNF
Any propositional formula can be equivalently transformed into the
Cnf:
1 Transform the formula into Nnf
2
Replace subformulas according to the table below, until Cnf
is obtained.
Rule
6
7
Subformula
(A ∧ B) ∨ C
C ∨ (A ∧ B)
Andrzej Szalas
Replaced by
(A ∨ C ) ∧ (B ∨ C )
(C ∨ A) ∧ (C ∨ B)
Slide 12 of 43
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Introduction & propositional calculus
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Lecture II: tableaux, satisfiability
Lecture III: resolution
Resolution rule for propositional calculus
Resolution method has been introduced by Robinson (1965) and
is considered the most powerful automated proving technique.
Resolution rule, denoted by (res), is formulated as follows:
α ∨ L,
¬L ∨ β
α∨β
Factorization rule for propositional calculus
Reminder
Factorization rule, denoted by (fctr ):
Remove from a clause redundant repetitions
of literals.
Example
P ∨ P ∨ ¬Q ∨ P ∨ ¬Q
P ∨ ¬Q
where L is a literal and α, β are clauses.
The position of L and ¬L in clauses does not matter.
Andrzej Szalas
Introduction & propositional calculus
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Slide 13 of 43
2
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Slide 14 of 43
Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
DeMorgan’s laws for quantifiers
The method, where formula A is to be proved
1
Andrzej Szalas
Lecture II: tableaux, satisfiability
Lecture III: resolution
Resolution method
Lecture II: tableaux, satisfiability
Lecture III: resolution
transform ¬A into the conjunctive normal form
try to derive the empty clause ∅ by applying resolution (res)
and factorization (fctr ):
if the empty clause is obtained then A is a tautology,
if the empty clause cannot be obtained no matter how (res)
and (fctr ) are applied, then conclude that A is not a tautology.
Andrzej Szalas
Slide 15 of 43
DeMorgan laws for quantifiers are:
[¬∀x A(x)] ↔ [∃x ¬A(x)]
[¬∃x A(x)] ↔ [∀x ¬A(x)]
Examples
1
“It is not the case that all animals are large”
is equivalent to “Some animals are not large”.
2
“It is not the case that some animals are plants”
is equivalent to “All animals are not plants”.
3
¬∀x∃y ∀z [friend(x, y ) ∧ likes(y , z)]
is equivalent to ∃x∀y ∃z ¬[friend(x, y ) ∧ likes(y , z)].
Andrzej Szalas
Slide 16 of 43
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Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
Other laws for quantifiers
1
Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
γ-formulas
Null quantification.
Assume variable x is not free in formula P. Then:
1
2
3
4
2
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Rule
(∀)
(¬∃)
∀x[P] is equivalent to P;
∃x[P] is equivalent to P;
∀x[P ∨ Q(x)] is equivalent to P ∨ ∀x[Q(x)];
∃x[P ∧ Q(x)] is equivalent to P ∧ ∃x[Q(x)].
2
γ(a)
A(a)
¬A(a)
where a is an arbitrary constant.
Pushing quantifiers past connectives:
1
γ
∀x A(x)
¬∃x A(x)
∀x[P(x) ∧ Q(x)] is equivalent to ∀x[P(x)] ∧ ∀x[Q(x)];
∃x[P(x) ∨ Q(x)] is equivalent to ∃x[P(x)] ∨ ∃x[Q(x)]
Andrzej Szalas
Introduction & propositional calculus
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Slide 17 of 43
Andrzej Szalas
Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
Introduction & propositional calculus
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Slide 18 of 43
Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
Tableaux
δ-formulas
δ-formulas
Rule
(∃)
(¬∀)
δ
∃x A(x)
¬∀x A(x)
δ(a)
A(a)
¬A(a)
where a is a fresh constant.
Construction of a semantic tableau for formula A
Initially T consists of a single unmarked node labeled with {A}.
Until possible choose anon-closed leaf l labeled with U(l) and
apply:
if U(l) contains a pair of complementary formulas then mark
the leaf closed;
otherwise:
Andrzej Szalas
Slide 19 of 43
Andrzej Szalas
Slide 20 of 43
Introduction & propositional calculus
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Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
Tableaux
Introduction & propositional calculus
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Transforming formulas into NNF
Construction of a semantic tableau for formula A
if U(l) does not contain a pair of complementary literals
and contains non-literals, choose a non-literal A ∈ U(l)
or a γ-formula from any node on the path from l
to the root, and:
if A is an α- or β-formula, apply rules provided in slide 7;
if A is a γ-formula, create a child node l ′ for l and label l ′
with U(l ′ ) = (U(l) − {A}) ∪ {γ(a)}, where a preferably
is a constant appearing in U(l);
if A is a δ-formula, create a child node l ′ for l and label l ′
with U(l ′ ) = (U(l) − {A}) ∪ {δ(a)}, where a is a constant
not appearing in U(l).
Andrzej Szalas
Introduction & propositional calculus
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NNF: Negation Normal Form
Recall (see Slide 9) that in negation normal form, Nnf, negation
ca appear only before atoms.
Any first-order formula can be equivalently transformed into the
Nnf by replacing subformulas according to the table given in
Slide 10 and rules given below.
Rule Subformula Replaced by
6’
¬∀x A(x)
∃x[¬A(x)]
7’
¬∃x A(x)
∀x[¬A(x)]
Slide 21 of 43
Andrzej Szalas
Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
PNF: Prenex Normal Form
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Q1 x1 Q2 x2 . . . Qn xn [A(x1 , x2 , . . . , xn )],
Examples
3
A(x) ∧ B(x, y ) ∧ ¬C (y ) is in Pnf
∀x∃y [A(x) ∧ B(x, y ) ∧ ¬C (y )] is in Pnf
A(x) ∧ ∀x [B(x, y ) ∧ ¬C (y )] as well as
∀x [A(x) ∧ B(x, y ) ∧ ¬∃yC (y )] are not in Pnf.
Andrzej Szalas
Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
Any predicate formula can be equivalently transformed into the
Pnf
1 Transform the formula into Nnf;
2
where n ≥ 0, Q1 , Q2 , . . . , Qn are quantifiers (∀, ∃) and A is
quantifier-free.
2
Slide 22 of 43
Transforming formulas into PNF
We say A is in prenex normal form, abbreviated by Pnf, if all its
quantifiers (if any) are in its prefix, i.e., it has the form:
1
Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
Slide 23 of 43
Replace subformulas according to the table below, until Pnf
is obtained, where Q denotes any quantifier, ∀ or ∃.
Rule
Subformula
12 Qx[A(x)], A(x) without variable z
13
∀x A(x) ∧ ∀x B(x)
14
∃x A(x) ∨ ∃x B(x)
15 A ∨ Qx B, where A contains no x
16 A ∧ Qx B, where A contains no x
Andrzej Szalas
Slide 24 of 43
Replaced by
Qz[A(z)]
∀x [A(x) ∧ B(x)]
∃x [A(x) ∨ B(x)]
Qx (A ∨ B)
Qx (A ∧ B)
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Lecture IV: introduction
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Skolem normal form
Transforming formulas into the Skolem normal form
Eliminate existential quantifiers from left to right:
1 when we have ∃xA(x) then remove ∃x and replace x in A
by a new constant symbol
when we have ∀x1 . . . ∀xk ∃xA(x1 , . . . , xk , x) then remove ∃x
and replace x in A by a term f (x1 , . . . , xk ), where f is a new
function symbol.
Andrzej Szalas
Introduction & propositional calculus
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Slide 25 of 43
Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
The unification algorithm
3
if the trees are identical then stop;
let t and t ′ be subtrees that have to be identical, but are not:
′
if t and t are function symbols/constants then conclude that
the substitutions do not exist and stop;
otherwise t or t ′ is a variable; let t be a variable, then
substitute all occurrences of t by the expression represented by
t ′ assuming that t does not occur in t ′ (if it occurs then
conclude that the substitutions do not exist and stop);
4
change the trees, according to the substitution determined in
the previous step and repeat from step 2.
Andrzej Szalas
Given two expressions, unification depends on substituting variables
by expressions so that both input expressions become identical.
If this is possible, the given expressions are said to be unifiable.
Examples
1
2
3
To unify expressions father (x) and father (mother (John)),
it suffices to substitute x by expression mother (John).
√
To unify expressions (x + f (y√)) and ( 2 ∗ z + f (3)),
it suffices to substitute x by 2 ∗ z and y by 3.
Expressions father (x) and mother (father (John))
cannot be unified.
Andrzej Szalas
Introduction & propositional calculus
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Slide 26 of 43
Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
Resolution rule for the predicate calculus
Input: expressions e, e ′
Output: substitution of variables which makes e and e ′ identical
or inform that such a substitution does not exist.
1 traverse trees corresponding to expressions e, e ′ ;
2
Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
Unification
A formula is in the Skolem normal form iff it is in the Pnf and
contains no existential quantifiers.
2
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Slide 27 of 43
Resolution rule, denoted by (res), is formulated for first-order
clauses as follows:
L1 (t¯1 ) ∨ . . . ∨ Lk−1 (t̄k−1 ) ∨ Lk (t̄k )
¬Lk (t¯k′ ) ∨ M1 (s̄1 ) ∨ . . . ∨ Ml (s̄l )
′
L1 (t¯1 ′ ) ∨ . . . ∨ Lk−1 (t̄k−1
) ∨ M1 (s̄1′ ) ∨ . . . ∨ Ml (s̄l′ )
,
where:
L1 , . . . , Lk , M1 , . . . , Ml are literals;
tk and tk′ are unifiable;
primed expressions are obtained from non-primed expressions
by applying substitutions unifying tk and tk′ .
Andrzej Szalas
Slide 28 of 43
Introduction & propositional calculus
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Lecture IV: introduction
Lecture V: tableaux
Lecture VI: resolution
Factorization rule for the predicate calculus
The rule
Factorization rule, denoted by (fctr ):
Unify some terms in a clause and remove from the
clause all repetitions of literals.
Examples
1
2
(fctr ) with x = Jack, y = mother (Eve):
parent(x, y ) ∨ parent(Jack, mother (Eve))
parent(Jack, mother (Eve))
(fctr ) with z = x, u = y :
P(x, y ) ∨ S(y , z, u) ∨ P(z, u)
P(x, y ) ∨ S(y , x, y )
Andrzej Szalas
Introduction & propositional calculus
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Introduction & propositional calculus
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Datalog
Facts in Datalog
Facts in Datalog are represented in the form of relations
name(arg1 , . . . , argk ), where name is a name of a relation
and arg1 , . . . , argk are constants.
Examples
1
address(John, ’Kungsgatan 12′ )
2
likes(John, Marc).
Slide 29 of 43
Lecture VII: deductive databases
Lecture VIII: sequent calculus
Atomic queries about facts
Lecture VII: deductive databases
Lecture VIII: sequent calculus
Andrzej Szalas
Introduction & propositional calculus
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Slide 30 of 43
Lecture VII: deductive databases
Lecture VIII: sequent calculus
Rules in Datalog
Atomic queries are of the form
name(arg1 , . . . , argk ),
where arg1 , . . . , argk are constants or variables, where ‘ ’ is also
a variable (without a name, “do-not-care” variable).
Rules in Datalog are expressed in the form of (a syntactic variant
of) Horn clauses:
R(Z̄ ):−R1 (Z̄1 ), . . . , Rk (Z̄k )
Examples
1
likes(John, Marc) – does John like Marc?
2
likes(X , Marc) – who likes Marc?
(compute X ’s satisfying likes(X , Marc))
3
likes(John, X ) – whom likes John?
4
likes(X , Y ) – compute all pairs hX , Y i such that likes(X , Y )
holds.
Andrzej Szalas
Slide 31 of 43
where Z̄ , Z̄1 , . . . , Z̄k are vectors of variable or constant symbols
such that any variable appearing on the lefthand side
of ‘:−’ (called the head of the rule) appears also at the righthand
side of the rule (called the body of the rule).
Andrzej Szalas
Slide 32 of 43
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Rules in Datalog
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Designing Datalog databases
Semantics
The intended meaning of the rule is that
[R1 (Z̄1 ) ∧ . . . ∧ Rk (Z̄k )] → R(Z̄ ),
where all variables that appear both in the rule’s head and body
are universally quantified, while those appearing only in the rule’s
body, are existentially quantified.
Remark
To define disjunction in rule’s body, one uses the following rules:
h:−b1.
h:−b2.
These rules are equivalent to
(b1 → h) ∧ (b2 → h)
Example
Rule: R(X , c):−Q(X , Z ), S(Z , X ) denotes implication:
∀X {∃Z [Q(X , Z ) ∧ S(Z , X )] → R(X , c)}.
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Introduction & propositional calculus
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Slide 33 of 43
Lecture VII: deductive databases
Lecture VIII: sequent calculus
Designing Datalog databases
which, in turn, is equivalent to: (b1 ∨ b2) → h.
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Introduction & propositional calculus
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Slide 34 of 43
Lecture VII: deductive databases
Lecture VIII: sequent calculus
The Gentzen system for propositional calculus
Remark
One also often uses the following tautology:
[(b → h1) ∧ (b → h2)] ↔ [b → (h1 ∧ h2)].
It allows to substitute:
h1 ∧ h2:−b
Gentzen systems are dual to tableaux or resolution in the sense that
we directly prove validity rather than unsatisfiability of formulas.
Sequents
By a sequent we understand any expression of the form
A ⇒ B, where A, B are finite sets of formulas.
Let A = {A1 , . . . , Ak } and B = {B1 , . . . , Bm }.
Then the sequent A ⇒ B represents:
A1 ∧ . . . ∧ Ak → B1 ∨ . . . ∨ Bm ,
by two rules:
h1:−b.
h2:−b.
where it is assumed that the empty disjunction is F and the
empty conjunction is T.
Andrzej Szalas
Slide 35 of 43
Andrzej Szalas
Slide 36 of 43
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Lecture VII: deductive databases
Lecture VIII: sequent calculus
Sequent-based rules and axioms
Lecture VII: deductive databases
Lecture VIII: sequent calculus
Rules for propositional connectives
Negation
General form of rules
A⇒B
− from A ⇒ B derive C ⇒ D
C ⇒D
A⇒B
from A ⇒ B derive conjunction
−
C ⇒ D and E ⇒ F
C ⇒ D; E ⇒ F
(¬l)
A, ¬B, C ⇒ D
A, C ⇒ B, D
Axioms
Axioms are of the form A, B, C ⇒ D, B, E ,
i.e., the same formula appears at both sides of ⇒.
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(∧r )
Slide 37 of 43
Lecture VII: deductive databases
Lecture VIII: sequent calculus
Rules for propositional connectives
A ⇒ B, ¬C , D
A, C ⇒ B, D
Conjunction
(∧l)
Andrzej Szalas
(¬r )
A, B ∧ C , D ⇒ E
A, B, C , D ⇒ E
A ⇒ B, C ∧ D, E
A ⇒ B, C , E ; A ⇒ B, D, E
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Introduction & propositional calculus
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Slide 38 of 43
Lecture VII: deductive databases
Lecture VIII: sequent calculus
Derived rules for definable connectives
Example: rule (→ l) for implication
Consider first A, B → C , D ⇒ E :
Disjunction
A, B ∨ C , D ⇒ E
(∨l)
A, B, D ⇒ E ; A, C , D ⇒ E
(∨r )
A ⇒ B, C ∨ D, E
A ⇒ B, C , D, E
(def )
A, B → C , D ⇒ E
A, ¬B ∨ C , D ⇒ E
(∨l)
A, ¬B, D ⇒ E
; A, C , D ⇒ E
(¬l)
A, D ⇒ B, E
Example: rule (→ l) for implication
Thus we have the following derived rule for implication:
(→ l)
Andrzej Szalas
Slide 39 of 43
A, B → C , D ⇒ E
A, D ⇒ B, E ; A, C , D ⇒ E
Andrzej Szalas
Slide 40 of 43
Introduction & propositional calculus
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Introduction & propositional calculus
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Lecture VII: deductive databases
Lecture VIII: sequent calculus
Derived rules for definable connectives
Lecture VII: deductive databases
Lecture VIII: sequent calculus
Gentzen rules for quantifiers
Example: rule (→ r )
Consider now A ⇒ B, C → D, E :
(def )
A ⇒ B, C → D, E
A ⇒ B, ¬C ∨ D, E
(∨r )
A ⇒ B, ¬C , D, E
(¬r )
A, C ⇒ B, D, E
Universal quantifier
(∀l)
A, ∀x B(x), C ⇒ D
A, ∀x B(x), C , B(a) ⇒ D
(∀r )
where, in application of (∀r ), a is a fresh constant.
Example: rule (→ r )
Thus we have the following derived rule:
(→ r )
A ⇒ B, C → D, E
A, C ⇒ B, D, E
Andrzej Szalas
Introduction & propositional calculus
First-order logic
Miscellaneous
Slide 41 of 43
Lecture VII: deductive databases
Lecture VIII: sequent calculus
Gentzen rules for quantifiers
Existential quantifier
(∃l)
A ⇒ B, ∀x C (x), D
A ⇒ B, C (a), D
A, ∃x B(x), C ⇒ D
A, B(a), C ⇒ D
(∃r )
A ⇒ B, ∃x C (x), D
A ⇒ B, ∃x C (x), D, C (a)
where, in application of (∃l), a is a fresh constant.
Andrzej Szalas
Slide 43 of 43
Andrzej Szalas
Slide 42 of 43