Termination of the given ITRSProblem could successfully be proven:



ITRS
  ↳ ITRStoIDPProof

ITRS problem:
The following domains are used:

z

The TRS R consists of the following rules:

rand(x, y) → Cond_rand(>@z(x, 0@z), x, y)
Cond_rand(TRUE, x, y) → rand(-@z(x, 1@z), id_inc(y))
rand(x, y) → Cond_rand1(=@z(x, 0@z), x, y)
id_inc(w(x)) → w(+@z(x, 1@z))
Cond_random(TRUE, x) → rand(x, w(0@z))
random(x) → Cond_random(>=@z(x, 0@z), x)
Cond_rand1(TRUE, x, y) → y
id_inc(w(x)) → w(x)

The set Q consists of the following terms:

rand(x0, x1)
Cond_rand(TRUE, x0, x1)
id_inc(w(x0))
Cond_random(TRUE, x0)
random(x0)
Cond_rand1(TRUE, x0, x1)


Added dependency pairs

↳ ITRS
  ↳ ITRStoIDPProof
IDP
      ↳ UsableRulesProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

rand(x, y) → Cond_rand(>@z(x, 0@z), x, y)
Cond_rand(TRUE, x, y) → rand(-@z(x, 1@z), id_inc(y))
rand(x, y) → Cond_rand1(=@z(x, 0@z), x, y)
id_inc(w(x)) → w(+@z(x, 1@z))
Cond_random(TRUE, x) → rand(x, w(0@z))
random(x) → Cond_random(>=@z(x, 0@z), x)
Cond_rand1(TRUE, x, y) → y
id_inc(w(x)) → w(x)

The integer pair graph contains the following rules and edges:

(0): RAND(x[0], y[0]) → COND_RAND1(=@z(x[0], 0@z), x[0], y[0])
(1): COND_RAND(TRUE, x[1], y[1]) → ID_INC(y[1])
(2): RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2])
(3): RANDOM(x[3]) → COND_RANDOM(>=@z(x[3], 0@z), x[3])
(4): COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4]))
(5): COND_RANDOM(TRUE, x[5]) → RAND(x[5], w(0@z))

(2) -> (1), if ((x[2]* x[1])∧(y[2]* y[1])∧(>@z(x[2], 0@z) →* TRUE))


(2) -> (4), if ((x[2]* x[4])∧(y[2]* y[4])∧(>@z(x[2], 0@z) →* TRUE))


(3) -> (5), if ((x[3]* x[5])∧(>=@z(x[3], 0@z) →* TRUE))


(4) -> (0), if ((id_inc(y[4]) →* y[0])∧(-@z(x[4], 1@z) →* x[0]))


(4) -> (2), if ((id_inc(y[4]) →* y[2])∧(-@z(x[4], 1@z) →* x[2]))


(5) -> (0), if ((x[5]* x[0]))


(5) -> (2), if ((x[5]* x[2]))



The set Q consists of the following terms:

rand(x0, x1)
Cond_rand(TRUE, x0, x1)
id_inc(w(x0))
Cond_random(TRUE, x0)
random(x0)
Cond_rand1(TRUE, x0, x1)


As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
IDP
          ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

id_inc(w(x)) → w(+@z(x, 1@z))
id_inc(w(x)) → w(x)

The integer pair graph contains the following rules and edges:

(0): RAND(x[0], y[0]) → COND_RAND1(=@z(x[0], 0@z), x[0], y[0])
(1): COND_RAND(TRUE, x[1], y[1]) → ID_INC(y[1])
(2): RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2])
(3): RANDOM(x[3]) → COND_RANDOM(>=@z(x[3], 0@z), x[3])
(4): COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4]))
(5): COND_RANDOM(TRUE, x[5]) → RAND(x[5], w(0@z))

(2) -> (1), if ((x[2]* x[1])∧(y[2]* y[1])∧(>@z(x[2], 0@z) →* TRUE))


(2) -> (4), if ((x[2]* x[4])∧(y[2]* y[4])∧(>@z(x[2], 0@z) →* TRUE))


(3) -> (5), if ((x[3]* x[5])∧(>=@z(x[3], 0@z) →* TRUE))


(4) -> (0), if ((id_inc(y[4]) →* y[0])∧(-@z(x[4], 1@z) →* x[0]))


(4) -> (2), if ((id_inc(y[4]) →* y[2])∧(-@z(x[4], 1@z) →* x[2]))


(5) -> (0), if ((x[5]* x[0]))


(5) -> (2), if ((x[5]* x[2]))



The set Q consists of the following terms:

rand(x0, x1)
Cond_rand(TRUE, x0, x1)
id_inc(w(x0))
Cond_random(TRUE, x0)
random(x0)
Cond_rand1(TRUE, x0, x1)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDependencyGraphProof
IDP
              ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

id_inc(w(x)) → w(+@z(x, 1@z))
id_inc(w(x)) → w(x)

The integer pair graph contains the following rules and edges:

(4): COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4]))
(2): RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2])

(2) -> (4), if ((x[2]* x[4])∧(y[2]* y[4])∧(>@z(x[2], 0@z) →* TRUE))


(4) -> (2), if ((id_inc(y[4]) →* y[2])∧(-@z(x[4], 1@z) →* x[2]))



The set Q consists of the following terms:

rand(x0, x1)
Cond_rand(TRUE, x0, x1)
id_inc(w(x0))
Cond_random(TRUE, x0)
random(x0)
Cond_rand1(TRUE, x0, x1)


The constraints were generated the following way:
The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4])) the following chains were created:




For Pair RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2]) the following chains were created:




To summarize, we get the following constraints P for the following pairs.



The constraints for P> respective Pbound are constructed from P where we just replace every occurence of "t ≥ s" in P by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(RAND(x1, x2)) = -1 + (2)x1   
POL(COND_RAND(x1, x2, x3)) = -1 + (2)x2   
POL(w(x1)) = -1 + (2)x1   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(id_inc(x1)) = -1 + (-1)x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4]))

The following pairs are in Pbound:

COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4]))

The following pairs are in P:

RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2])

At least the following rules have been oriented under context sensitive arithmetic replacement:

-@z1
+@z1


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDependencyGraphProof
            ↳ IDP
              ↳ IDPNonInfProof
IDP
                  ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

id_inc(w(x)) → w(+@z(x, 1@z))
id_inc(w(x)) → w(x)

The integer pair graph contains the following rules and edges:

(2): RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2])


The set Q consists of the following terms:

rand(x0, x1)
Cond_rand(TRUE, x0, x1)
id_inc(w(x0))
Cond_random(TRUE, x0)
random(x0)
Cond_rand1(TRUE, x0, x1)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.