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:

cond(TRUE, x, y) → +@z(1@z, minus(x, +@z(y, 1@z)))
min(u, v) → if(<@z(u, v), u, v)
if(FALSE, u, v) → v
if(TRUE, u, v) → u
minus(x, x) → 0@z
minus(x, y) → cond(=@z(min(x, y), y), x, y)

The set Q consists of the following terms:

cond(TRUE, x0, x1)
min(x0, x1)
if(FALSE, x0, x1)
if(TRUE, x0, x1)
minus(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:

cond(TRUE, x, y) → +@z(1@z, minus(x, +@z(y, 1@z)))
min(u, v) → if(<@z(u, v), u, v)
if(FALSE, u, v) → v
if(TRUE, u, v) → u
minus(x, x) → 0@z
minus(x, y) → cond(=@z(min(x, y), y), x, y)

The integer pair graph contains the following rules and edges:

(0): MINUS(x[0], y[0]) → MIN(x[0], y[0])
(1): MINUS(x[1], y[1]) → COND(=@z(min(x[1], y[1]), y[1]), x[1], y[1])
(2): MIN(u[2], v[2]) → IF(<@z(u[2], v[2]), u[2], v[2])
(3): COND(TRUE, x[3], y[3]) → MINUS(x[3], +@z(y[3], 1@z))

(0) -> (2), if ((y[0]* v[2])∧(x[0]* u[2]))


(1) -> (3), if ((x[1]* x[3])∧(y[1]* y[3])∧(=@z(min(x[1], y[1]), y[1]) →* TRUE))


(3) -> (0), if ((+@z(y[3], 1@z) →* y[0])∧(x[3]* x[0]))


(3) -> (1), if ((+@z(y[3], 1@z) →* y[1])∧(x[3]* x[1]))



The set Q consists of the following terms:

cond(TRUE, x0, x1)
min(x0, x1)
if(FALSE, x0, x1)
if(TRUE, x0, x1)
minus(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:

min(u, v) → if(<@z(u, v), u, v)
if(FALSE, u, v) → v
if(TRUE, u, v) → u

The integer pair graph contains the following rules and edges:

(0): MINUS(x[0], y[0]) → MIN(x[0], y[0])
(1): MINUS(x[1], y[1]) → COND(=@z(min(x[1], y[1]), y[1]), x[1], y[1])
(2): MIN(u[2], v[2]) → IF(<@z(u[2], v[2]), u[2], v[2])
(3): COND(TRUE, x[3], y[3]) → MINUS(x[3], +@z(y[3], 1@z))

(0) -> (2), if ((y[0]* v[2])∧(x[0]* u[2]))


(1) -> (3), if ((x[1]* x[3])∧(y[1]* y[3])∧(=@z(min(x[1], y[1]), y[1]) →* TRUE))


(3) -> (0), if ((+@z(y[3], 1@z) →* y[0])∧(x[3]* x[0]))


(3) -> (1), if ((+@z(y[3], 1@z) →* y[1])∧(x[3]* x[1]))



The set Q consists of the following terms:

cond(TRUE, x0, x1)
min(x0, x1)
if(FALSE, x0, x1)
if(TRUE, x0, x1)
minus(x0, x1)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 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:

min(u, v) → if(<@z(u, v), u, v)
if(FALSE, u, v) → v
if(TRUE, u, v) → u

The integer pair graph contains the following rules and edges:

(1): MINUS(x[1], y[1]) → COND(=@z(min(x[1], y[1]), y[1]), x[1], y[1])
(3): COND(TRUE, x[3], y[3]) → MINUS(x[3], +@z(y[3], 1@z))

(1) -> (3), if ((x[1]* x[3])∧(y[1]* y[3])∧(=@z(min(x[1], y[1]), y[1]) →* TRUE))


(3) -> (1), if ((+@z(y[3], 1@z) →* y[1])∧(x[3]* x[1]))



The set Q consists of the following terms:

cond(TRUE, x0, x1)
min(x0, x1)
if(FALSE, x0, x1)
if(TRUE, x0, x1)
minus(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 MINUS(x[1], y[1]) → COND(=@z(min(x[1], y[1]), y[1]), x[1], y[1]) the following chains were created:




For Pair COND(TRUE, x[3], y[3]) → MINUS(x[3], +@z(y[3], 1@z)) 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)) = -1   
POL(if(x1, x2, x3)) = (-1)x3 + (-1)x2 + (-1)x1   
POL(TRUE) = -1   
POL(MINUS(x1, x2)) = -1 + (-1)x2 + x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = -1   
POL(COND(x1, x2, x3)) = -1 + (-1)x3 + x2   
POL(<@z(x1, x2)) = -1   
POL(min(x1, x2)) = 1 + x2 + x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND(TRUE, x[3], y[3]) → MINUS(x[3], +@z(y[3], 1@z))

The following pairs are in Pbound:

COND(TRUE, x[3], y[3]) → MINUS(x[3], +@z(y[3], 1@z))

The following pairs are in P:

MINUS(x[1], y[1]) → COND(=@z(min(x[1], y[1]), y[1]), x[1], y[1])

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

+@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:

min(u, v) → if(<@z(u, v), u, v)
if(FALSE, u, v) → v
if(TRUE, u, v) → u

The integer pair graph contains the following rules and edges:

(1): MINUS(x[1], y[1]) → COND(=@z(min(x[1], y[1]), y[1]), x[1], y[1])


The set Q consists of the following terms:

cond(TRUE, x0, x1)
min(x0, x1)
if(FALSE, x0, x1)
if(TRUE, x0, x1)
minus(x0, x1)


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