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:

cond1(TRUE, x, y) → +@z(y, 1@z)
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
ackNat(FALSE, x, y) → 0@z
f(TRUE, x) → f(>=@z(ack(10@z, 10@z), x), +@z(x, 1@z))
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, 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:

cond1(TRUE, x, y) → +@z(y, 1@z)
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
ackNat(FALSE, x, y) → 0@z
f(TRUE, x) → f(>=@z(ack(10@z, 10@z), x), +@z(x, 1@z))
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(0): F(TRUE, x[0]) → ACK(10@z, 10@z)
(1): ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])
(2): COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])
(3): COND2(TRUE, x[3], y[3]) → ACK(-@z(x[3], 1@z), +@z(y[3], 1@z))
(4): ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])
(5): F(TRUE, x[5]) → F(>=@z(ack(10@z, 10@z), x[5]), +@z(x[5], 1@z))
(6): COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))
(7): COND2(FALSE, x[7], y[7]) → ACK(-@z(x[7], 1@z), ack(x[7], -@z(y[7], 1@z)))

(0) -> (1), if (ACK(10@z, 10@z) →* ACK(x[1], y[1]))


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


(2) -> (3), if ((x[2] →* x[3])∧(y[2] →* y[3])∧(=@z(y[2], 0@z) →* TRUE))


(2) -> (6), if ((x[2] →* x[6])∧(y[2] →* y[6])∧(=@z(y[2], 0@z) →* FALSE))


(2) -> (7), if ((x[2] →* x[7])∧(y[2] →* y[7])∧(=@z(y[2], 0@z) →* FALSE))


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


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


(5) -> (0), if ((+@z(x[5], 1@z) →* x[0])∧(>=@z(ack(10@z, 10@z), x[5]) →* TRUE))


(5) -> (5), if ((+@z(x[5], 1@z) →* x[5]a)∧(>=@z(ack(10@z, 10@z), x[5]) →* TRUE))


(6) -> (1), if ((-@z(y[6], 1@z) →* y[1])∧(x[6] →* x[1]))


(7) -> (1), if ((ack(x[7], -@z(y[7], 1@z)) →* y[1])∧(-@z(x[7], 1@z) →* x[1]))



The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, 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:

cond1(TRUE, x, y) → +@z(y, 1@z)
ackNat(FALSE, x, y) → 0@z
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(0): F(TRUE, x[0]) → ACK(10@z, 10@z)
(1): ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])
(2): COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])
(3): COND2(TRUE, x[3], y[3]) → ACK(-@z(x[3], 1@z), +@z(y[3], 1@z))
(4): ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])
(5): F(TRUE, x[5]) → F(>=@z(ack(10@z, 10@z), x[5]), +@z(x[5], 1@z))
(6): COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))
(7): COND2(FALSE, x[7], y[7]) → ACK(-@z(x[7], 1@z), ack(x[7], -@z(y[7], 1@z)))

(0) -> (1), if (ACK(10@z, 10@z) →* ACK(x[1], y[1]))


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


(2) -> (3), if ((x[2] →* x[3])∧(y[2] →* y[3])∧(=@z(y[2], 0@z) →* TRUE))


(2) -> (6), if ((x[2] →* x[6])∧(y[2] →* y[6])∧(=@z(y[2], 0@z) →* FALSE))


(2) -> (7), if ((x[2] →* x[7])∧(y[2] →* y[7])∧(=@z(y[2], 0@z) →* FALSE))


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


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


(5) -> (0), if ((+@z(x[5], 1@z) →* x[0])∧(>=@z(ack(10@z, 10@z), x[5]) →* TRUE))


(5) -> (5), if ((+@z(x[5], 1@z) →* x[5]a)∧(>=@z(ack(10@z, 10@z), x[5]) →* TRUE))


(6) -> (1), if ((-@z(y[6], 1@z) →* y[1])∧(x[6] →* x[1]))


(7) -> (1), if ((ack(x[7], -@z(y[7], 1@z)) →* y[1])∧(-@z(x[7], 1@z) →* x[1]))



The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, x0, x1)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

cond1(TRUE, x, y) → +@z(y, 1@z)
ackNat(FALSE, x, y) → 0@z
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(3): COND2(TRUE, x[3], y[3]) → ACK(-@z(x[3], 1@z), +@z(y[3], 1@z))
(7): COND2(FALSE, x[7], y[7]) → ACK(-@z(x[7], 1@z), ack(x[7], -@z(y[7], 1@z)))
(2): COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])
(4): ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])
(1): ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])
(6): COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))

(2) -> (3), if ((x[2] →* x[3])∧(y[2] →* y[3])∧(=@z(y[2], 0@z) →* TRUE))


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


(6) -> (1), if ((-@z(y[6], 1@z) →* y[1])∧(x[6] →* x[1]))


(7) -> (1), if ((ack(x[7], -@z(y[7], 1@z)) →* y[1])∧(-@z(x[7], 1@z) →* x[1]))


(2) -> (6), if ((x[2] →* x[6])∧(y[2] →* y[6])∧(=@z(y[2], 0@z) →* FALSE))


(2) -> (7), if ((x[2] →* x[7])∧(y[2] →* y[7])∧(=@z(y[2], 0@z) →* FALSE))


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


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



The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, 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 COND2(TRUE, x[3], y[3]) → ACK(-@z(x[3], 1@z), +@z(y[3], 1@z)) the following chains were created:




For Pair COND2(FALSE, x[7], y[7]) → ACK(-@z(x[7], 1@z), ack(x[7], -@z(y[7], 1@z))) the following chains were created:




For Pair COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2]) the following chains were created:




For Pair ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4]) the following chains were created:




For Pair ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1]) the following chains were created:




For Pair COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 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)) = x1 + (-1)x2   
POL(ackNat(x1, x2, x3)) = -1 + (-1)x3 + (2)x2 + (-1)x1   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND2(x1, x2, x3)) = -1 + x2   
POL(cond2(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2   
POL(FALSE) = -1   
POL(=@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(cond1(x1, x2, x3)) = -1 + (2)x2 + (-1)x1   
POL(ACKNAT(x1, x2, x3)) = -1 + x2   
POL(ACK(x1, x2)) = -1 + x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(COND1(x1, x2, x3)) = -1 + x2   
POL(ack(x1, x2)) = -1 + x2 + (-1)x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND2(TRUE, x[3], y[3]) → ACK(-@z(x[3], 1@z), +@z(y[3], 1@z))
COND2(FALSE, x[7], y[7]) → ACK(-@z(x[7], 1@z), ack(x[7], -@z(y[7], 1@z)))

The following pairs are in Pbound:

COND2(TRUE, x[3], y[3]) → ACK(-@z(x[3], 1@z), +@z(y[3], 1@z))

The following pairs are in P≥:

COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])
ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])
ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])
COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))

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

&&(FALSE, FALSE)1 ↔ FALSE1
-@z1 ↔
&&(TRUE, TRUE)1 ↔ TRUE1
+@z1 ↔
&&(TRUE, FALSE)1 → FALSE1
&&(FALSE, TRUE)1 → FALSE1


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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

cond1(TRUE, x, y) → +@z(y, 1@z)
ackNat(FALSE, x, y) → 0@z
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(7): COND2(FALSE, x[7], y[7]) → ACK(-@z(x[7], 1@z), ack(x[7], -@z(y[7], 1@z)))
(2): COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])
(4): ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])
(1): ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])
(6): COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))

(6) -> (1), if ((-@z(y[6], 1@z) →* y[1])∧(x[6] →* x[1]))


(7) -> (1), if ((ack(x[7], -@z(y[7], 1@z)) →* y[1])∧(-@z(x[7], 1@z) →* x[1]))


(2) -> (6), if ((x[2] →* x[6])∧(y[2] →* y[6])∧(=@z(y[2], 0@z) →* FALSE))


(2) -> (7), if ((x[2] →* x[7])∧(y[2] →* y[7])∧(=@z(y[2], 0@z) →* FALSE))


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


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



The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, 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 COND2(FALSE, x[7], y[7]) → ACK(-@z(x[7], 1@z), ack(x[7], -@z(y[7], 1@z))) the following chains were created:




For Pair COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2]) the following chains were created:




For Pair ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4]) the following chains were created:




For Pair ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1]) the following chains were created:




For Pair COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 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)) = x1 + (-1)x2   
POL(ackNat(x1, x2, x3)) = 1 + (-1)x3 + x1   
POL(0@z) = 0   
POL(TRUE) = 1   
POL(&&(x1, x2)) = 1   
POL(COND2(x1, x2, x3)) = x2   
POL(cond2(x1, x2, x3)) = -1 + (-1)x3 + x2 + (-1)x1   
POL(FALSE) = 1   
POL(=@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(cond1(x1, x2, x3)) = -1 + (-1)x3 + (2)x2 + x1   
POL(ACKNAT(x1, x2, x3)) = 1 + x2 + (-1)x1   
POL(ACK(x1, x2)) = x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(COND1(x1, x2, x3)) = x2   
POL(ack(x1, x2)) = 2 + (2)x2 + (-1)x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND2(FALSE, x[7], y[7]) → ACK(-@z(x[7], 1@z), ack(x[7], -@z(y[7], 1@z)))

The following pairs are in Pbound:

COND2(FALSE, x[7], y[7]) → ACK(-@z(x[7], 1@z), ack(x[7], -@z(y[7], 1@z)))

The following pairs are in P≥:

COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])
ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])
ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])
COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))

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

&&(FALSE, FALSE)1 ↔ FALSE1
-@z1 ↔
+@z1 ↔
&&(TRUE, TRUE)1 ↔ TRUE1
&&(FALSE, TRUE)1 ↔ FALSE1
&&(TRUE, FALSE)1 ↔ FALSE1


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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

cond1(TRUE, x, y) → +@z(y, 1@z)
ackNat(FALSE, x, y) → 0@z
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(2): COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])
(4): ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])
(1): ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])
(6): COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))

(6) -> (1), if ((-@z(y[6], 1@z) →* y[1])∧(x[6] →* x[1]))


(2) -> (6), if ((x[2] →* x[6])∧(y[2] →* y[6])∧(=@z(y[2], 0@z) →* FALSE))


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


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



The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, 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 COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2]) the following chains were created:




For Pair ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4]) the following chains were created:




For Pair ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1]) the following chains were created:




For Pair COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 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)) = x1 + (-1)x2   
POL(ackNat(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (-1)x1   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND2(x1, x2, x3)) = -1 + (2)x3   
POL(cond2(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (-1)x1   
POL(FALSE) = -1   
POL(=@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(cond1(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (-1)x1   
POL(ACKNAT(x1, x2, x3)) = -1 + (2)x3 + (-1)x1   
POL(ACK(x1, x2)) = (2)x2   
POL(+@z(x1, x2)) = x1 + x2   
POL(COND1(x1, x2, x3)) = -1 + (2)x3   
POL(ack(x1, x2)) = -1 + (-1)x2 + (-1)x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])

The following pairs are in Pbound:

COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))

The following pairs are in P≥:

COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])
ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])
COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))

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

&&(FALSE, FALSE)1 ↔ FALSE1
-@z1 ↔
TRUE1 → &&(TRUE, TRUE)1
FALSE1 → &&(TRUE, FALSE)1
&&(FALSE, TRUE)1 ↔ FALSE1


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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

cond1(TRUE, x, y) → +@z(y, 1@z)
ackNat(FALSE, x, y) → 0@z
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(2): COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])
(4): ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])
(1): ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])

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


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



The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, x0, x1)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

cond1(TRUE, x, y) → +@z(y, 1@z)
ackNat(FALSE, x, y) → 0@z
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(2): COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])
(1): ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])
(6): COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))

(6) -> (1), if ((-@z(y[6], 1@z) →* y[1])∧(x[6] →* x[1]))


(2) -> (6), if ((x[2] →* x[6])∧(y[2] →* y[6])∧(=@z(y[2], 0@z) →* FALSE))



The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, x0, x1)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

cond1(TRUE, x, y) → +@z(y, 1@z)
ackNat(FALSE, x, y) → 0@z
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(2): COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])
(4): ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])
(1): ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])
(6): COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))

(6) -> (1), if ((-@z(y[6], 1@z) →* y[1])∧(x[6] →* x[1]))


(2) -> (6), if ((x[2] →* x[6])∧(y[2] →* y[6])∧(=@z(y[2], 0@z) →* FALSE))


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


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



The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, 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 COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2]) the following chains were created:




For Pair ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4]) the following chains were created:




For Pair ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1]) the following chains were created:




For Pair COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 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)) = x1 + (-1)x2   
POL(ackNat(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (-1)x1   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND2(x1, x2, x3)) = -1 + (2)x3   
POL(cond2(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (-1)x1   
POL(FALSE) = -1   
POL(=@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(cond1(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (-1)x1   
POL(ACKNAT(x1, x2, x3)) = -1 + (2)x3 + (-1)x1   
POL(ACK(x1, x2)) = (2)x2   
POL(+@z(x1, x2)) = x1 + x2   
POL(COND1(x1, x2, x3)) = (2)x3   
POL(ack(x1, x2)) = -1 + (-1)x2 + (-1)x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])

The following pairs are in Pbound:

COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))

The following pairs are in P≥:

ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])
ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])
COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))

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

FALSE1 → &&(FALSE, FALSE)1
-@z1 ↔
TRUE1 → &&(TRUE, TRUE)1
&&(TRUE, FALSE)1 ↔ FALSE1
FALSE1 → &&(FALSE, TRUE)1


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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

cond1(TRUE, x, y) → +@z(y, 1@z)
ackNat(FALSE, x, y) → 0@z
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(2): COND1(FALSE, x[2], y[2]) → COND2(=@z(y[2], 0@z), x[2], y[2])
(4): ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])
(1): ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])

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


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



The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, x0, x1)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

cond1(TRUE, x, y) → +@z(y, 1@z)
ackNat(FALSE, x, y) → 0@z
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(4): ACKNAT(TRUE, x[4], y[4]) → COND1(=@z(x[4], 0@z), x[4], y[4])
(1): ACK(x[1], y[1]) → ACKNAT(&&(>=@z(x[1], 0@z), >=@z(y[1], 0@z)), x[1], y[1])
(6): COND2(FALSE, x[6], y[6]) → ACK(x[6], -@z(y[6], 1@z))

(6) -> (1), if ((-@z(y[6], 1@z) →* y[1])∧(x[6] →* x[1]))


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



The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, x0, x1)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

cond1(TRUE, x, y) → +@z(y, 1@z)
ackNat(FALSE, x, y) → 0@z
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(5): F(TRUE, x[5]) → F(>=@z(ack(10@z, 10@z), x[5]), +@z(x[5], 1@z))

(5) -> (5), if ((+@z(x[5], 1@z) →* x[5]a)∧(>=@z(ack(10@z, 10@z), x[5]) →* TRUE))



The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, 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 F(TRUE, x[5]) → F(>=@z(ack(10@z, 10@z), x[5]), +@z(x[5], 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)) = x1 + (-1)x2   
POL(ackNat(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (2)x1   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(cond2(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (2)x1   
POL(FALSE) = -1   
POL(=@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(cond1(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (2)x1   
POL(10@z) = 10   
POL(+@z(x1, x2)) = x1 + x2   
POL(F(x1, x2)) = -1 + (-1)x2   
POL(ack(x1, x2)) = -1 + (2)x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

F(TRUE, x[5]) → F(>=@z(ack(10@z, 10@z), x[5]), +@z(x[5], 1@z))

The following pairs are in Pbound:

F(TRUE, x[5]) → F(>=@z(ack(10@z, 10@z), x[5]), +@z(x[5], 1@z))

The following pairs are in P≥:
none

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

+@z1 ↔


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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

cond1(TRUE, x, y) → +@z(y, 1@z)
ackNat(FALSE, x, y) → 0@z
cond2(TRUE, x, y) → ack(-@z(x, 1@z), +@z(y, 1@z))
ack(x, y) → ackNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
cond1(FALSE, x, y) → cond2(=@z(y, 0@z), x, y)
ackNat(TRUE, x, y) → cond1(=@z(x, 0@z), x, y)
cond2(FALSE, x, y) → ack(-@z(x, 1@z), ack(x, -@z(y, 1@z)))

The integer pair graph is empty.
The set Q consists of the following terms:

cond1(TRUE, x0, x1)
ack(x0, x1)
ackNat(FALSE, x0, x1)
f(TRUE, x0)
cond2(TRUE, x0, x1)
cond1(FALSE, x0, x1)
ackNat(TRUE, x0, x1)
cond2(FALSE, x0, x1)


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