1) Metodom rezolucije dokazati:
 
  H = (Ax)(Ay)(p(x,y)=>p(y,x))
  K = (Ax)(Ay)(Az)((p(x,y)/\p(y,z)) => p(x,z))
  L = (Ax)((Ey)p(x,y) => p(x,x))

 Dokazati: (H /\ K) => L

H /\ K /\ ~L nezadovoljiva

(Ax)(Ay)(p(x,y)=>p(y,x))
(Ax)(Ay)(Az)((p(x,y)/\p(y,z)) => p(x,z))
(Ex)((Ey)p(x,y) /\ ~p(x,x))

NNF:

(Ax)(Ay)(~p(x,y) \/ p(y,x))                  /\
(Ax)(Ay)(Az)(~p(x,y)\/~p(y,z) \/ p(x,z))     /\
(Ex)((Ey)p(x,y) /\ ~p(x,x))

PRENEX:

(Ex)(Ev)(Au)(Ay)(Az)
    (~p(u,y) \/ p(y,u))              /\
    (~p(u,y)\/~p(y,z) \/ p(u,z))     /\
    (p(x,v) /\ ~p(x,x))

Skolemizacija

(Au)(Ay)(Az)
    (~p(u,y) \/ p(y,u))              /\
    (~p(u,y)\/~p(y,z) \/ p(u,z))     /\
    (p(a,b) /\ ~p(a,a))


CNF:
1) ~p(u,y), p(y,u)
2)  ~p(u,y), ~p(y,z), p(u,z)
3) p(a,b)
4) ~p(a,a)

Rezolucija:
5) p(b, a)             1), 3) u-->a, y --> b
6) ~p(b, z), p(a, z)   2), 3) u->a, y->b
7) p(a,a)              5), 6) z-->a
8) []                  4), 7)



2) Metodom rezolucije dokazati:

  P = (Ax)((s(x) /\ t(x)) => r(x)) => (Ex)(s(x) /\ ~t(x))
  Q = (Ax)(s(x) => t(x)) \/ (Ax)(s(x) => r(x))
  R = (Ax)((s(x) /\ r(x)) => t(x)) => (Ex)(s(x)/\t(x)/\~r(x))
 
 Dokazati (P /\ Q) => R

 Dokazujemo da je P /\ Q /\ ~R nezadovoljiva

(Ax)((s(x) /\ t(x)) => r(x)) => (Ex)(s(x) /\ ~t(x))
(Ax)(s(x) => t(x)) \/ (Ax)(s(x) => r(x))
(Ax)((s(x) /\ r(x)) => t(x)) /\ (Ax)(~s(x)\/~t(x)\/r(x))

NNF:

(Ex)(s(x) /\ t(x) /\ ~r(x)) \/ (Ex)(s(x) /\ ~t(x))
(Ax)(~s(x) \/ t(x)) \/ (Ax)(~s(x) \/ r(x))
(Ax)(~s(x) \/ ~r(x) \/ t(x)) /\ (Ax)(~s(x)\/~t(x)\/r(x))

PRENEX:

(Ex)(Av)(Au)
    ((s(x) /\ t(x) /\ ~r(x)) \/ (s(x) /\ ~t(x)))
    ((~s(v) \/ t(v)) \/ (~s(u) \/ r(u)))
    ((~s(v) \/ ~r(v) \/ t(v)) /\ (~s(v)\/~t(v)\/r(v)))

SKOLEMIZACIJA:

   (Av)(Au)
    ((s(c) /\ t(c) /\ ~r(c)) \/ (s(c) /\ ~t(c)))
    ((~s(v) \/ t(v)) \/ (~s(u) \/ r(u)))
    ((~s(v) \/ ~r(v) \/ t(v)) /\ (~s(v)\/~t(v)\/r(v)))

CNF:
1) s(c)
2) s(c), ~t(c)
3) t(c), s(c)
4) ~r(c), s(c)
5) ~r(c), ~t(c)
6) ~s(v), t(v), ~s(u), r(u)
7) ~s(v),~r(v),t(v)
8) ~s(v), ~t(v), r(v)

Rezolucija:
9) t(c), r(c)          (1, 6) v --> c, u-->c
10) ~r(c),t(c)         (1, 7) v--> c
11) ~t(c), r(c)         (1,8)  v-->c
12) ~t(c)               (11, 5)
13) t(c)                (9,10)
14) []                  (12, 13)


3) Dokazati da je sledeca formula valjana:

((Ax)(p(x) => q(x)) /\
(Ax)(q(x) => s(x)) /\
(Ax)(r(x) => s(x)) /\
(Ax)(p(x) \/ r(x)))      => (Ax)s(x)

Negacija: 

((Ax)(p(x) => q(x)) /\
(Ax)(q(x) => s(x)) /\
(Ax)(r(x) => s(x)) /\
(Ax)(p(x) \/ r(x)))     /\ ~(Ax)s(x)

NNF:

((Ax)(~p(x) \/ q(x)) /\
(Ax)(~q(x) \/ s(x)) /\
(Ax)(~r(x) \/ s(x)) /\
(Ax)(p(x) \/ r(x)))     /\ (Ex)~s(x)

PRENEX:
(Ex)(Ay)
(~p(y) \/ q(y)) /\
(~q(y) \/ s(y)) /\
(~r(y) \/ s(y)) /\
(p(y) \/ r(y))) /\ ~s(x)

Skolemizacija:

(Ay)
(~p(y) \/ q(y)) /\
(~q(y) \/ s(y)) /\
(~r(y) \/ s(y)) /\
(p(y) \/ r(y))) /\ ~s(c)

CNF:

1) (~p(y) \/ q(y)) 
2) (~q(y) \/ s(y)) 
3) (~r(y) \/ s(y)) 
4) (p(y) \/ r(y)))
5) ~s(c)

Rezolucija:

6) ~q(c)   (2, 5)
7) ~r(c)   (3, 5)
8) ~p(c)   (1, 6)
9) p(c)    (4, 7)
10) []     (8,9)

PARAMODULACIJA:

C1 \/ t = s    C2[t']
----------------------
(C1 \/ C2[s])Sigma     , Sigma je n.o.u za t i t'


4) Dokazati da je sledeca formula valjana u jednakosnoj logici:

(Ax)(Ay)(Az)(f(f(x,y),z) = f(x,f(y,z))) /\
(Ax)(n(x) <=> (Ay)(f(x,y) = y /\ f(y,x) = y)) =>
(Ax)(Ay)(n(x) /\ n(y) => x = y)

Negacija:

(Ax)(Ay)(Az)(f(f(x,y),z) = f(x,f(y,z))) /\
(Ax)(n(x) <=> (Ay)(f(x,y) = y /\ f(y,x) = y)) /\
(Ex)(Ey)(n(x) /\ n(y) /\ x != y)

NNF:

(Ax)(Ay)(Az)(f(f(x,y),z) = f(x,f(y,z))) /\
(Ax)(~n(x) \/ (Ay)(f(x,y) = y /\ f(y,x) = y) /\
      n(x)  \/ (Ey)(f(x,y) != y \/ f(y,x) != y)) /\
(Ex)(Ey)(n(x) /\ n(y) /\ x != y)

PRENEX:

(Ex)(Ey)(Au)(Ev)(Aw)(Az)

(f(f(u,w),z) = f(u,f(w,z))) /\
(~n(u) \/ (f(u,w) = w /\ f(w,u) = w) /\
n(u)  \/ (f(u,v) != v \/ f(v,u) != v)) /\
(n(x) /\ n(y) /\ x != y)

SKOLEMIZACIJA:

(Au)(Aw)(Az)
(f(f(u,w),z) = f(u,f(w,z))) /\
(~n(u) \/ (f(u,w) = w /\ f(w,u) = w) /\
n(u)  \/ (f(u,g(u)) != g(u) \/ f(g(u),u) != g(u))) /\
(n(a) /\ n(b) /\ a != b)

CNF:
----

1) f(f(u,w),z) = f(u,f(w,z))
2) ~n(u), f(u,w) = w
3) ~n(u), f(w,u) = w
4) n(u)  \/ f(u,g(u)) != g(u) \/ f(g(u),u) != g(u)
5) n(a)
6) n(b)
7) a != b

Rezolucija + paramodulacija

8) f(a, w) = w     (2, 5)
9) f(w, a) = w     (3, 5)
10) f(b, w) = w    (2, 6)
11) f(w, b) = w    (3, 6)
12) a = b          (Paramodulacija, 11, 8)
13) []             (7, 12)

INTUITIVNI DOKAZ:
a = f(a, b) = b
a = b

5) Dokazati da je sledeca formula valjana u jednakosnoj logici:

(Ax)(Ay)(Az)(f(f(x,y),z) = f(x,f(y,z))) /\
(Ax)(f(e,x) = x /\ f(x,e) = x) /\
(Ax)(Ay)(i(x, y) <=> f(x,y) = e /\ f(y,x) = e)  =>
(Ax)(Ay)(Az)(i(x,y) /\ i(x,z) => y = z)

Negacija:

(Ax)(Ay)(Az)(f(f(x,y),z) = f(x,f(y,z))) /\
(Ax)(f(e,x) = x /\ f(x,e) = x) /\
(Ax)(Ay)(i(x, y) <=> f(x,y) = e /\ f(y,x) = e) /\
(Ex)(Ey)(Ez)(i(x,y) /\ i(x,z) /\ y != z)

NNF:

(Ax)(Ay)(Az)(f(f(x,y),z) = f(x,f(y,z))) /\
(Ax)(f(e,x) = x /\ f(x,e) = x) /\
(Ax)(Ay)(~i(x, y) \/ (f(x,y) = e /\ f(y,x) = e) /\
         i(x, y) \/ f(x,y) != e \/ f(y,x) != e   ) /\
(Ex)(Ey)(Ez)(i(x,y) /\ i(x,z) /\ y != z)

PRENEX:

(Ex)(Ey)(Ez)(Au)(Av)(Aw)

(f(f(u,v),w) = f(u,f(v,w))) /\
(f(e,u) = u /\ f(u,e) = u) /\
(~i(u, v) \/ (f(u,v) = e /\ f(v,u) = e) /\
        i(u, v) \/ f(u,v) != e \/ f(v,u) != e   ) /\
(i(x,y) /\ i(x,z) /\ y != z)


Skolemizacija:

(Au)(Av)(Aw)
(f(f(u,v),w) = f(u,f(v,w))) /\
(f(e,u) = u /\ f(u,e) = u) /\
(~i(u, v) \/ (f(u,v) = e /\ f(v,u) = e) /\
        i(u, v) \/ f(u,v) != e \/ f(v,u) != e   ) /\
(i(a,b) /\ i(a,c) /\ b != c)

CNF:

1) f(f(u,v),w) = f(u,f(v,w))
2) f(e,u) = u
3) f(u,e) = u
4) ~i(u, v) \/ f(u,v) = e
5) ~i(u, v) \/ f(v,u) = e
6) i(u, v) \/ f(u,v) != e \/ f(v,u) != e
7) i(a,b)
8) i(a,c)
9) b != c

Rezolucija:

10) f(a, b) = e  4,7
11) f(b, a) = e  5,7
12) f(a, c) = e  4,8
13) f(c, a) = e  5,8
14) f(e,w) = f(c, f(a, w))   Paramodulacija: 13 u 1
15) f(e,b) = f(c, e)         Paramodulacija: 10 u 14
16) b = f(c,e)               Paramodulacija: 2 u 15
17) b = c                    Paramodulacija: 3 u 16
18) []                       9, 17

INTUITIVNI DOKAZ:
(b =  f(e,b) = f(f(c,a),b) = f(c,f(a,b)) = f(c,e)= c)
(b =  f(e,b) = f(c,f(a,b)) = f(c,e)= c)
(b =  f(e,b) = f(c,e)= c)
(b = f(c,e)= c)
(b = c)
