Propositional logic
Problem 1
1) Using the truth table method check if “  ” connective is associative:
p  (q  r )  ( p  q)  r .
2) Using the truth table method check if “  ” connective is associative:
p  (q  r )  ( p  q)  r .
3) Using the truth table method check if “  ” connective is distributive over “  ”
connective: p  (q  r )  ( p  q)  ( p  r ) .
4) Using the truth table method check if “  ” connective is distributive over “  ”
connective: p  (q  r )  ( p  q)  ( p  r ) .
5) Using the truth table method check the properties of absorption: p  (q  p)  p and
p  (q  p)  p .
6) Using the truth table method check the De’Morgan law for  :
( p  q)  p  q ;
Problem 2.
Using the truth table method decide what kind of formula (consistent, inconsistent,
tautology, contingent) is A . If A is consistent, write all the models of A.
1) A= q  p  r  p  (q  r )
2) A= p  (q  r )  q  p
3) A= p  (q  r )  q  p  r
4) A= (p  q)  r  p  (q  r )
5) A= p  (q  r )  q  p
6)
A  p  (q  r)  q  p  r .
Problem 3. Using the truth table method check if:
1) p  q |= ( p  r )  ( p  q  r )
2) p  q |= (q  r )  ( p  r )
3) p  (q  r ) |= ( p  q)  ( p  r )
4) p  r |= (q  r )  (( p  q)  r )
5) p  q |= (p  q)  q
6)
p  q | (q  r)  ( p  q  r) ;
Problem 4.
Using the truth table method prove that the following formulas are tautologies.
1) ( p  (q  r ))  (( p  r )  (q  r )) --- semidistributivity of  over 
2) ( p  (q  r ))  (q  ( p  r )) --- permutation of the premises law.
3) ( p  (q  r ))  ( p  q  r ) --- reunion of the premises law.
4) ( p  q  r )  ( p  (q  r )) --- separation of the premises law.
5) ( p  q)  ( p  q  r )  ( p  r ) ---“cut” law.
6) p  (q  r)  (( p  q)  ( p  r)) --- semidistributivity of  over 
Problem 5. Transform the formula A into its equivalent CNF and DNF. Using one of
these forms prove that A is a valid formula in propositional calculus.
1) A= ( p  (q  r ))  (q  ( p  r )) --- permutation of the premises law.
2) A= ( p  q)  ( p  q  r )  ( p  r ) ---“cut” law.
3) A= ( p  q  r )  ( p  (q  r )) --- separation of the premises law.
4) A= ( p  (q  r ))  ( p  q  r ) --- reunion of the premises law.
5) A= (U  (V  Z ))  ((U  V )  (U  Z )) --- axiom A2
6) A= ( p  (q  r ))  (( p  r )  (q  r )) --- semidistributivity of  over 
Problem 6.
Using the apropriate normal form write all the models of the following formulas:
1) A= ( p  q  r )  ( p  r )  q .
2) A= ( p  q  r )  ( p  r )  q
3) A= (q  r  p)  ( p  r )  q
4) A= ( p  q  r )  (q  r )  p
5) A= (q  r  p)  ( p  r )  q
6) A= (p  q)  r  p  (q  r)
Problem 7 .
Using the apropriate normal form prove that the following formulas are inconsistent:
1)
2)
3)
4)
5)
6)
7)
8)
(U  (V  Z ))  ((U  V )  (U  Z )) ;
(U V )  (V  U ) ;
(U V )  (U V  Z )  (U  Z ) ;
(U  (V  Z ))  ((U  V )  (U  Z )) ;
U  (V  Z )  ((U V )  (U  Z )) ;
(U  (V  Z ))  (U V  Z ) ;
(U  (V  Z ))  (V  (U  Z )) ;
(U V  Z )  (U  (V  Z )) ;
Problem 8
Prove the following properties of the logical equivqlence relation where: R, S  FP and
U, V, Z  FP.
1. monotonicity: if R |=U and R  S then S |=U;
2. cut: if S |=Vi, iI and S{Vi|iI }|= U then S |=U;
3. tranzitivity: if S |=U and {U}|=V then S |=V;
4. conjunction in conclusions (right „and” ):
if S |= U and S |= V then S |= UV;
5. disjunction in premises (left „or”):
if S{U} |= Z and S{V} |= Z then S{UV}|=Z;
6. proof by cases:
if S{U} |= V and S{U } |= V then S |= V.
Problem 9.
Using the definition of deduction prove the following:
1. p  q, r  p, r | q ;
2. p  r, p  r  q, r | q ;
3. p  q, q  r, p | r ;
4. p  (q  r), p  q, p | r ;
5. p  (q  r), q, p | r ;
6. p  (q  r), p  q, p | r ;
7. p  (q  r), p  q, p | r ;
8. p  q, r  t, p  r, q | t .
Problem 10.
Using the theorem of deduction and its reverse prove that:
1) |  (U  (V  Z )  (U  V  Z ) --- reunion of the premises law.
2) |  (U  V  Z )  (U  (V  Z )) --- separation of the premises law
3) |  (U  (V  Z ))  (V  (U  Z )) --- permutation of the premises law.
4) |  (U  (V  Z ))  ((U  V )  (U  Z )) --- second axiom of propositional logic
5) | ( p  q)  ( p  q  r )  ( p  r ) --- “cut” law.
6) | ( p  q)  (( p  r)  ( p  q  r))
Problem 11.
Using the theorem of deduction and its reverse prove that:
1) |  ( p  q)  (( r  p)  (r  q))
2) |  ( p  r )  (( q  r )  ( p  q  r ))
3) |  ( p  q)  (( r  t )  ( p  r  q  t ))
4) |  ( p  r )  (( p  r  q)  ( p  q))
5) |  (q  p)  ((s  q)  (s  p))
6)
| (U  (V  Z ))  (U  (Z  V )) .
Problem 12
Using the theorem of deduction and its reverse prove that:
1) |  ( p  q)  (( p  r )  ( p  q  r ))
2) |  ( p  (q  r ))  (q  ( p  r ))
3) |  ( p  (q  r ))  (( p  q)  ( p  r ))
4) |  ( p  q)  ( p  q  r )  ( p  r )
5)
6)
|  q  (( p  (q  r ))  ( p  r ))
| p  (q  r )  (( p  q)  ( p  r )) ;
Problem 13. Using the semantic tableaux method decide what kind of formula is A. If A is consistent,
write all its models.
1. A= (( p  q)  (p  r ))  (q  r ) .
2. A= ( p  q  r )  ( p  r )  q
3. A= (q  r  p)  ( p  r )  q
4. A= (( r  q)  (p  r ))  ( p  q) .
5. A= (( p  r )  (p  r ))  (q  r )
6. A  ( ( p  r)  p)  (( p  q)  ( p  r)) .
Problem 14. Prove that the formula A is valid using the semantic tableaux method.
1.A= ( p  (q  r ))  (q  ( p  r )) permutation of the premises
2. A= ( p  q  r )  ( p  q)  ( p  r ) distributivity of implication over disjunction
3. A= ( p  q  r )  ( p  q)  ( p  r ) distributivity of implication over disjunction
4. A= ( p  q)  (( r  t )  ( p  r  q  t ))
5. A= ((V  Z )  U )  ((U V )  (U  Z ))
6. A  ( p  (q  r))  ( p  q  r) ;
Problem 15. Using the semantic tableaux method check if:
1. p  (q  r  s), p, s | q .
2. p  (q  r ), r  q |= (p  q)  r
3. p  (q  r  s), p, r | q
4. p  q, r  t , p  r | q  t
5. p  (q  r ), q  r |= p  (q  r )
Problem 16. Using the sequent calculus check if the following sequent is true or not.
1. p  q  p  r , r  q  r .
2.
3.
4.
5.
6.
p  q  r  ( p  r)  q
q  r  p, q  ( p  r )  q
p, r  q  p  r  ( p  q)  (q  p) .
p  r , p  q  q  r , r  q
p  (q  r), q  r | p  (q  r)
Problem 17. Prove the validity of the formula A using the sequent calculus method:
1) A= ( p  q  r )  ( p  q)  ( p  r ) , distributivity of  over 
2) A= ( p  (q  r ))  ( p  q  r ) , reunion of the premises
3) A= ((V  Z )  U )  ((U  V )  (U  Z ))
4) A= ( p  q  r )  ( p  q)  ( p  r ) , distributivity of  over  .
5) A= ( p  q  r )  ( p  (q  r )) , separation of the premises
6) A  p  (q  r)  ( p  q)  ( p  r)
Problem 18. Using the sequent calculus check if:
1) p  (q  r ), r  q |= (p  q)  r
2)
3)
4)
5)
6)
p  (q  r ), q  r |= p  (q  r )
p  q, r  t , p  r | q  t
p  (q  r  s), p, s | q  s
p  (q  r  s), p, r | q  p
p  (q  r  s), p, r | q  r
Problem 19.
Using general resolution prove that the following formulas are tautologies:
1) (B  A)  ((B  A)  B)
2) ( B  A)  (C  A)  ( B  C  A)
3) ( A  B)  ((A  C )  (B  C ))
4) ( A  B)  ((C  A)  (C  B))
5) ( p  q  r )  ( p  q)  ( p  r )
Problem 20
Using lock resolution check the inconsistency of the following sets of clauses:
1) { p  q, p  q  r , p  q  r , p  r , p  r} ,
2) { p  q, p  q  r , p  r , p  r , p  r} ,
3) { p  q, p  q  r , p  q  r , q  r , q  r} ,
4) { p  q, p  q  r , p  q  r , p  q, p  r} ,
5) { p  r , p  q  r , p  q  r , p  q  r , p  q  r , r} ,
Problem 21
Check if the following deductions are valid using linear resolution:
1)
2)
3)
4)
5)
p  q  r , p  q  r , q  r |  r
p  q  r , p  q, q  r |  p
p  q  r , p  q  r , q  p |  q
p  q, q  r , p  r |  r
p  q  r , p  q  r , p  r |  p