Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), x, y, exp(y, z))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), x, y, exp(y, z))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PLUS(s(x), y) → PLUS(x, y)
TIMES(s(x), y) → PLUS(y, times(x, y))
TIMES(s(x), y) → TIMES(x, y)
EXP(x, s(y)) → TIMES(x, exp(x, y))
EXP(x, s(y)) → EXP(x, y)
GE(s(x), s(y)) → GE(x, y)
TOWER(x, y) → TOWERITER(0, x, y, s(0))
TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z)
TOWERITER(c, x, y, z) → GE(c, x)
HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z))
HELP(false, c, x, y, z) → EXP(y, z)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), x, y, exp(y, z))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), x, y, exp(y, z))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(8) UsableRulesProof (EQUIVALENT transformation)

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.

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

GE(s(x), s(y)) → GE(x, y)

R is empty.
The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

GE(s(x), s(y)) → GE(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• GE(s(x), s(y)) → GE(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

(13) TRUE

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), x, y, exp(y, z))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(15) UsableRulesProof (EQUIVALENT transformation)

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.

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PLUS(s(x), y) → PLUS(x, y)

R is empty.
The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PLUS(s(x), y) → PLUS(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• PLUS(s(x), y) → PLUS(x, y)
  The graph contains the following edges 1 > 1, 2 >= 2

(20) TRUE

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), x, y, exp(y, z))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(22) UsableRulesProof (EQUIVALENT transformation)

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.

(23) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TIMES(s(x), y) → TIMES(x, y)

R is empty.
The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TIMES(s(x), y) → TIMES(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• TIMES(s(x), y) → TIMES(x, y)
  The graph contains the following edges 1 > 1, 2 >= 2

(27) TRUE

(28) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EXP(x, s(y)) → EXP(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), x, y, exp(y, z))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(29) UsableRulesProof (EQUIVALENT transformation)

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.

(30) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EXP(x, s(y)) → EXP(x, y)

R is empty.
The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(31) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

(32) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EXP(x, s(y)) → EXP(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(33) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• EXP(x, s(y)) → EXP(x, y)
  The graph contains the following edges 1 >= 1, 2 > 2

(34) TRUE

(35) Obligation:

Q DP problem:
The TRS P consists of the following rules:

HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z))
TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), x, y, exp(y, z))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(36) UsableRulesProof (EQUIVALENT transformation)

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.

(37) Obligation:

Q DP problem:
The TRS P consists of the following rules:

HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z))
TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, x) → x
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(38) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

tower(x0, x1)
towerIter(x0, x1, x2, x3)
help(true, x0, x1, x2, x3)
help(false, x0, x1, x2, x3)

(39) Obligation:

Q DP problem:
The TRS P consists of the following rules:

HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z))
TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, x) → x
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(40) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z) we obtained the following new rules [LPAR04]:

TOWERITER(s(z0), z1, z2, y_0) → HELP(ge(s(z0), z1), s(z0), z1, z2, y_0)

(41) Obligation:

Q DP problem:
The TRS P consists of the following rules:

HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z))
TOWERITER(s(z0), z1, z2, y_0) → HELP(ge(s(z0), z1), s(z0), z1, z2, y_0)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, x) → x
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(42) NonInfProof (EQUIVALENT transformation)

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 HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z)) the following chains were created:

• We consider the chain TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z), HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z)) which results in the following constraint:
  

  (1)    (HELP(ge(x4, x5), x4, x5, x6, x7)=HELP(false, x8, x9, x10, x11) ⇒ HELP(false, x8, x9, x10, x11)≥TOWERITER(s(x8), x9, x10, exp(x10, x11)))

  
  
  We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
  

  (2)    (ge(x4, x5)=false ⇒ HELP(false, x4, x5, x6, x7)≥TOWERITER(s(x4), x5, x6, exp(x6, x7)))

  
  
  We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x4, x5)=false which results in the following new constraints:
  

  (3)    (false=false ⇒ HELP(false, 0, s(x25), x6, x7)≥TOWERITER(s(0), s(x25), x6, exp(x6, x7)))

  
  

  (4)    (ge(x27, x26)=false∧(∀x28,x29:ge(x27, x26)=false ⇒ HELP(false, x27, x26, x28, x29)≥TOWERITER(s(x27), x26, x28, exp(x28, x29))) ⇒ HELP(false, s(x27), s(x26), x6, x7)≥TOWERITER(s(s(x27)), s(x26), x6, exp(x6, x7)))

  
  
  We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
  

  (5)    (HELP(false, 0, s(x25), x6, x7)≥TOWERITER(s(0), s(x25), x6, exp(x6, x7)))

  
  
  We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x28,x29:ge(x27, x26)=false ⇒ HELP(false, x27, x26, x28, x29)≥TOWERITER(s(x27), x26, x28, exp(x28, x29))) with σ = [x28 / x6, x29 / x7] which results in the following new constraint:
  

  (6)    (HELP(false, x27, x26, x6, x7)≥TOWERITER(s(x27), x26, x6, exp(x6, x7)) ⇒ HELP(false, s(x27), s(x26), x6, x7)≥TOWERITER(s(s(x27)), s(x26), x6, exp(x6, x7)))

  
  
  

For Pair TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z) the following chains were created:

• We consider the chain HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z)), TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z) which results in the following constraint:
  

  (7)    (TOWERITER(s(x12), x13, x14, exp(x14, x15))=TOWERITER(x16, x17, x18, x19) ⇒ TOWERITER(x16, x17, x18, x19)≥HELP(ge(x16, x17), x16, x17, x18, x19))

  
  
  We simplified constraint (7) using rules (I), (II), (III), (IV) which results in the following new constraint:
  

  (8)    (TOWERITER(s(x12), x13, x14, x19)≥HELP(ge(s(x12), x13), s(x12), x13, x14, x19))

  
  
  

To summarize, we get the following constraints P_≥ for the following pairs.

• HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z))
  
  • (HELP(false, 0, s(x25), x6, x7)≥TOWERITER(s(0), s(x25), x6, exp(x6, x7)))
    
  • (HELP(false, x27, x26, x6, x7)≥TOWERITER(s(x27), x26, x6, exp(x6, x7)) ⇒ HELP(false, s(x27), s(x26), x6, x7)≥TOWERITER(s(s(x27)), s(x26), x6, exp(x6, x7)))
    
  
  
• TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z)
  
  • (TOWERITER(s(x12), x13, x14, x19)≥HELP(ge(s(x12), x13), s(x12), x13, x14, x19))
    
  
  

The constraints for P_> respective P_bound 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 P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(HELP(x₁, x₂, x₃, x₄, x₅)) = -1 - x₁ - x₂ + x₃ + x₄   
POL(TOWERITER(x₁, x₂, x₃, x₄)) = -1 - x₁ + x₂ + x₃   
POL(c) = -2   
POL(exp(x₁, x₂)) = 0   
POL(false) = 0   
POL(ge(x₁, x₂)) = 0   
POL(plus(x₁, x₂)) = 0   
POL(s(x₁)) = 1 + x₁   
POL(times(x₁, x₂)) = x₂   
POL(true) = 0   

The following pairs are in P_>:

HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z))

The following pairs are in P_bound:

HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z))

The following rules are usable:

false → ge(0, s(x))
ge(x, y) → ge(s(x), s(y))
true → ge(x, 0)

(43) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, x) → x
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
exp(x0, 0)
exp(x0, s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(44) DependencyGraphProof (EQUIVALENT transformation)

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

(45) TRUE