YES Termination w.r.t. Q proof of AProVE_07_otto12.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QReductionProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPSizeChangeProof (⇔, 0 ms)
        ↳27 YES
    ↳28 QDP
        ↳29 UsableRulesProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 QReductionProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QDPSizeChangeProof (⇔, 0 ms)
        ↳34 YES
    ↳35 QDP
        ↳36 UsableRulesProof (⇔, 0 ms)
        ↳37 QDP
        ↳38 QReductionProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 NonInfProof (⇔, 37 ms)
        ↳41 AND
            ↳42 QDP
                ↳43 DependencyGraphProof (⇔, 0 ms)
                ↳44 TRUE
            ↳45 QDP
                ↳46 DependencyGraphProof (⇔, 0 ms)
                ↳47 TRUE

(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) YES

(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) YES

(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) YES

(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) YES

(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) 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(x4, x5, x6, x7) → HELP(ge(x4, x5), x4, x5, x6, x7), HELP(false, x8, x9, x10, x11) → TOWERITER(s(x8), x9, x10, exp(x10, x11)) 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, x12, x13, x14, x15) → TOWERITER(s(x12), x13, x14, exp(x14, x15)), TOWERITER(x16, x17, x18, x19) → HELP(ge(x16, x17), x16, x17, x18, x19) which results in the following constraint:
    (1)    (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 (1) using rules (I), (II), (III), (IV) which results in the following new constraint:
    (2)    (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 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 [NONINF]:

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

The following pairs are in P>:

TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z)
The following pairs are in Pbound:

HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z))
The following rules are usable:

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

(41) Complex Obligation (AND)

(42) 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))

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.

(43) DependencyGraphProof (EQUIVALENT transformation)

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

(44) TRUE

(45) 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.

(46) DependencyGraphProof (EQUIVALENT transformation)

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

(47) TRUE