YES Termination w.r.t. Q proof of GTSSK07_cade15.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 (⇔, 49 ms)
        ↳41 QDP
        ↳42 DependencyGraphProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 UsableRulesProof (⇔, 0 ms)
        ↳45 QDP
        ↳46 QReductionProof (⇔, 0 ms)
        ↳47 QDP
        ↳48 NonInfProof (⇔, 12 ms)
        ↳49 QDP
        ↳50 DependencyGraphProof (⇔, 0 ms)
        ↳51 TRUE

(0) Obligation:

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

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

(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:

SORT(l) → ST(0, l)
ST(n, l) → COND1(member(n, l), n, l)
ST(n, l) → MEMBER(n, l)
COND1(true, n, l) → ST(s(n), l)
COND1(false, n, l) → COND2(gt(n, max(l)), n, l)
COND1(false, n, l) → GT(n, max(l))
COND1(false, n, l) → MAX(l)
COND2(false, n, l) → ST(s(n), l)
MEMBER(n, cons(m, l)) → OR(equal(n, m), member(n, l))
MEMBER(n, cons(m, l)) → EQUAL(n, m)
MEMBER(n, cons(m, l)) → MEMBER(n, l)
EQUAL(s(x), s(y)) → EQUAL(x, y)
GT(s(u), s(v)) → GT(u, v)
MAX(cons(u, l)) → IF(gt(u, max(l)), u, max(l))
MAX(cons(u, l)) → GT(u, max(l))
MAX(cons(u, l)) → MAX(l)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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:

GT(s(u), s(v)) → GT(u, v)

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

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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].

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

(11) Obligation:

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

GT(s(u), s(v)) → GT(u, v)

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:

  • GT(s(u), s(v)) → GT(u, v)
    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:

MAX(cons(u, l)) → MAX(l)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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:

MAX(cons(u, l)) → MAX(l)

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

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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].

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

(18) Obligation:

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

MAX(cons(u, l)) → MAX(l)

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:

  • MAX(cons(u, l)) → MAX(l)
    The graph contains the following edges 1 > 1

(20) YES

(21) Obligation:

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

EQUAL(s(x), s(y)) → EQUAL(x, y)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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:

EQUAL(s(x), s(y)) → EQUAL(x, y)

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

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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].

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

(25) Obligation:

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

EQUAL(s(x), s(y)) → EQUAL(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:

  • EQUAL(s(x), s(y)) → EQUAL(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:

MEMBER(n, cons(m, l)) → MEMBER(n, l)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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:

MEMBER(n, cons(m, l)) → MEMBER(n, l)

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

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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].

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

(32) Obligation:

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

MEMBER(n, cons(m, l)) → MEMBER(n, l)

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:

  • MEMBER(n, cons(m, l)) → MEMBER(n, l)
    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:

COND1(true, n, l) → ST(s(n), l)
ST(n, l) → COND1(member(n, l), n, l)
COND1(false, n, l) → COND2(gt(n, max(l)), n, l)
COND2(false, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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:

COND1(true, n, l) → ST(s(n), l)
ST(n, l) → COND1(member(n, l), n, l)
COND1(false, n, l) → COND2(gt(n, max(l)), n, l)
COND2(false, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
if(true, u, v) → u
if(false, u, v) → v
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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].

sort(x0)
st(x0, x1)
cond1(true, x0, x1)
cond1(false, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)

(39) Obligation:

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

COND1(true, n, l) → ST(s(n), l)
ST(n, l) → COND1(member(n, l), n, l)
COND1(false, n, l) → COND2(gt(n, max(l)), n, l)
COND2(false, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
if(true, u, v) → u
if(false, u, v) → v
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, 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 COND1(true, n, l) → ST(s(n), l) the following chains were created:
  • We consider the chain ST(x2, x3) → COND1(member(x2, x3), x2, x3), COND1(true, x4, x5) → ST(s(x4), x5) which results in the following constraint:
    (1)    (COND1(member(x2, x3), x2, x3)=COND1(true, x4, x5) ⇒ COND1(true, x4, x5)≥ST(s(x4), x5))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (member(x2, x3)=true ⇒ COND1(true, x2, x3)≥ST(s(x2), x3))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on member(x2, x3)=true which results in the following new constraint:
    (3)    (or(equal(x45, x44), member(x45, x43))=true∧(member(x45, x43)=true ⇒ COND1(true, x45, x43)≥ST(s(x45), x43)) ⇒ COND1(true, x45, cons(x44, x43))≥ST(s(x45), cons(x44, x43)))


    We simplified constraint (3) using rule (VII) which results in the following new constraint:
    (4)    (equal(x45, x44)=x46∧member(x45, x43)=x47∧or(x46, x47)=true∧(member(x45, x43)=true ⇒ COND1(true, x45, x43)≥ST(s(x45), x43)) ⇒ COND1(true, x45, cons(x44, x43))≥ST(s(x45), cons(x44, x43)))


    We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on or(x46, x47)=true which results in the following new constraints:
    (5)    (true=true∧equal(x45, x44)=x48∧member(x45, x43)=true∧(member(x45, x43)=true ⇒ COND1(true, x45, x43)≥ST(s(x45), x43)) ⇒ COND1(true, x45, cons(x44, x43))≥ST(s(x45), cons(x44, x43)))

    (6)    (x49=true∧equal(x45, x44)=x49∧member(x45, x43)=false∧(member(x45, x43)=true ⇒ COND1(true, x45, x43)≥ST(s(x45), x43)) ⇒ COND1(true, x45, cons(x44, x43))≥ST(s(x45), cons(x44, x43)))


    We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
    (7)    (equal(x45, x44)=x48∧member(x45, x43)=true∧(member(x45, x43)=true ⇒ COND1(true, x45, x43)≥ST(s(x45), x43)) ⇒ COND1(true, x45, cons(x44, x43))≥ST(s(x45), cons(x44, x43)))


    We simplified constraint (6) using rules (III), (IV) which results in the following new constraint:
    (8)    (equal(x45, x44)=true ⇒ COND1(true, x45, cons(x44, x43))≥ST(s(x45), cons(x44, x43)))


    We simplified constraint (7) using rule (VI) where we applied the induction hypothesis (member(x45, x43)=true ⇒ COND1(true, x45, x43)≥ST(s(x45), x43)) with σ = [ ] which results in the following new constraint:
    (9)    (equal(x45, x44)=x48∧COND1(true, x45, x43)≥ST(s(x45), x43) ⇒ COND1(true, x45, cons(x44, x43))≥ST(s(x45), cons(x44, x43)))


    We simplified constraint (9) using rule (IV) which results in the following new constraint:
    (10)    (COND1(true, x45, x43)≥ST(s(x45), x43) ⇒ COND1(true, x45, cons(x44, x43))≥ST(s(x45), cons(x44, x43)))


    We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on equal(x45, x44)=true which results in the following new constraints:
    (11)    (true=true ⇒ COND1(true, 0, cons(0, x43))≥ST(s(0), cons(0, x43)))

    (12)    (equal(x53, x52)=true∧(∀x54:equal(x53, x52)=true ⇒ COND1(true, x53, cons(x52, x54))≥ST(s(x53), cons(x52, x54))) ⇒ COND1(true, s(x53), cons(s(x52), x43))≥ST(s(s(x53)), cons(s(x52), x43)))


    We simplified constraint (11) using rules (I), (II) which results in the following new constraint:
    (13)    (COND1(true, 0, cons(0, x43))≥ST(s(0), cons(0, x43)))


    We simplified constraint (12) using rule (VI) where we applied the induction hypothesis (∀x54:equal(x53, x52)=true ⇒ COND1(true, x53, cons(x52, x54))≥ST(s(x53), cons(x52, x54))) with σ = [x54 / x43] which results in the following new constraint:
    (14)    (COND1(true, x53, cons(x52, x43))≥ST(s(x53), cons(x52, x43)) ⇒ COND1(true, s(x53), cons(s(x52), x43))≥ST(s(s(x53)), cons(s(x52), x43)))






For Pair ST(n, l) → COND1(member(n, l), n, l) the following chains were created:
  • We consider the chain COND1(true, x10, x11) → ST(s(x10), x11), ST(x12, x13) → COND1(member(x12, x13), x12, x13) which results in the following constraint:
    (1)    (ST(s(x10), x11)=ST(x12, x13) ⇒ ST(x12, x13)≥COND1(member(x12, x13), x12, x13))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (ST(s(x10), x11)≥COND1(member(s(x10), x11), s(x10), x11))


  • We consider the chain COND2(false, x18, x19) → ST(s(x18), x19), ST(x20, x21) → COND1(member(x20, x21), x20, x21) which results in the following constraint:
    (1)    (ST(s(x18), x19)=ST(x20, x21) ⇒ ST(x20, x21)≥COND1(member(x20, x21), x20, x21))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (ST(s(x18), x19)≥COND1(member(s(x18), x19), s(x18), x19))






For Pair COND1(false, n, l) → COND2(gt(n, max(l)), n, l) the following chains were created:
  • We consider the chain ST(x24, x25) → COND1(member(x24, x25), x24, x25), COND1(false, x26, x27) → COND2(gt(x26, max(x27)), x26, x27) which results in the following constraint:
    (1)    (COND1(member(x24, x25), x24, x25)=COND1(false, x26, x27) ⇒ COND1(false, x26, x27)≥COND2(gt(x26, max(x27)), x26, x27))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (member(x24, x25)=false ⇒ COND1(false, x24, x25)≥COND2(gt(x24, max(x25)), x24, x25))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on member(x24, x25)=false which results in the following new constraints:
    (3)    (false=false ⇒ COND1(false, x55, nil)≥COND2(gt(x55, max(nil)), x55, nil))

    (4)    (or(equal(x58, x57), member(x58, x56))=false∧(member(x58, x56)=false ⇒ COND1(false, x58, x56)≥COND2(gt(x58, max(x56)), x58, x56)) ⇒ COND1(false, x58, cons(x57, x56))≥COND2(gt(x58, max(cons(x57, x56))), x58, cons(x57, x56)))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (COND1(false, x55, nil)≥COND2(gt(x55, max(nil)), x55, nil))


    We simplified constraint (4) using rule (VII) which results in the following new constraint:
    (6)    (equal(x58, x57)=x59∧member(x58, x56)=x60∧or(x59, x60)=false∧(member(x58, x56)=false ⇒ COND1(false, x58, x56)≥COND2(gt(x58, max(x56)), x58, x56)) ⇒ COND1(false, x58, cons(x57, x56))≥COND2(gt(x58, max(cons(x57, x56))), x58, cons(x57, x56)))


    We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on or(x59, x60)=false which results in the following new constraint:
    (7)    (x62=false∧equal(x58, x57)=x62∧member(x58, x56)=false∧(member(x58, x56)=false ⇒ COND1(false, x58, x56)≥COND2(gt(x58, max(x56)), x58, x56)) ⇒ COND1(false, x58, cons(x57, x56))≥COND2(gt(x58, max(cons(x57, x56))), x58, cons(x57, x56)))


    We simplified constraint (7) using rule (VI) where we applied the induction hypothesis (member(x58, x56)=false ⇒ COND1(false, x58, x56)≥COND2(gt(x58, max(x56)), x58, x56)) with σ = [ ] which results in the following new constraint:
    (8)    (x62=false∧equal(x58, x57)=x62∧COND1(false, x58, x56)≥COND2(gt(x58, max(x56)), x58, x56) ⇒ COND1(false, x58, cons(x57, x56))≥COND2(gt(x58, max(cons(x57, x56))), x58, cons(x57, x56)))


    We simplified constraint (8) using rule (III) which results in the following new constraint:
    (9)    (equal(x58, x57)=false∧COND1(false, x58, x56)≥COND2(gt(x58, max(x56)), x58, x56) ⇒ COND1(false, x58, cons(x57, x56))≥COND2(gt(x58, max(cons(x57, x56))), x58, cons(x57, x56)))


    We simplified constraint (9) using rule (V) (with possible (I) afterwards) using induction on equal(x58, x57)=false which results in the following new constraints:
    (10)    (false=false∧COND1(false, s(x63), x56)≥COND2(gt(s(x63), max(x56)), s(x63), x56) ⇒ COND1(false, s(x63), cons(0, x56))≥COND2(gt(s(x63), max(cons(0, x56))), s(x63), cons(0, x56)))

    (11)    (false=false∧COND1(false, 0, x56)≥COND2(gt(0, max(x56)), 0, x56) ⇒ COND1(false, 0, cons(s(x64), x56))≥COND2(gt(0, max(cons(s(x64), x56))), 0, cons(s(x64), x56)))

    (12)    (equal(x66, x65)=false∧COND1(false, s(x66), x56)≥COND2(gt(s(x66), max(x56)), s(x66), x56)∧(∀x67:equal(x66, x65)=false∧COND1(false, x66, x67)≥COND2(gt(x66, max(x67)), x66, x67) ⇒ COND1(false, x66, cons(x65, x67))≥COND2(gt(x66, max(cons(x65, x67))), x66, cons(x65, x67))) ⇒ COND1(false, s(x66), cons(s(x65), x56))≥COND2(gt(s(x66), max(cons(s(x65), x56))), s(x66), cons(s(x65), x56)))


    We simplified constraint (10) using rules (I), (II) which results in the following new constraint:
    (13)    (COND1(false, s(x63), x56)≥COND2(gt(s(x63), max(x56)), s(x63), x56) ⇒ COND1(false, s(x63), cons(0, x56))≥COND2(gt(s(x63), max(cons(0, x56))), s(x63), cons(0, x56)))


    We simplified constraint (11) using rules (I), (II) which results in the following new constraint:
    (14)    (COND1(false, 0, x56)≥COND2(gt(0, max(x56)), 0, x56) ⇒ COND1(false, 0, cons(s(x64), x56))≥COND2(gt(0, max(cons(s(x64), x56))), 0, cons(s(x64), x56)))


    We simplified constraint (12) using rule (IV) which results in the following new constraint:
    (15)    (COND1(false, s(x66), x56)≥COND2(gt(s(x66), max(x56)), s(x66), x56) ⇒ COND1(false, s(x66), cons(s(x65), x56))≥COND2(gt(s(x66), max(cons(s(x65), x56))), s(x66), cons(s(x65), x56)))






For Pair COND2(false, n, l) → ST(s(n), l) the following chains were created:
  • We consider the chain COND1(false, x36, x37) → COND2(gt(x36, max(x37)), x36, x37), COND2(false, x38, x39) → ST(s(x38), x39) which results in the following constraint:
    (1)    (COND2(gt(x36, max(x37)), x36, x37)=COND2(false, x38, x39) ⇒ COND2(false, x38, x39)≥ST(s(x38), x39))


    We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
    (2)    (max(x37)=x68∧gt(x36, x68)=false ⇒ COND2(false, x36, x37)≥ST(s(x36), x37))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x36, x68)=false which results in the following new constraints:
    (3)    (false=false∧max(x37)=x69 ⇒ COND2(false, 0, x37)≥ST(s(0), x37))

    (4)    (gt(x72, x71)=false∧max(x37)=s(x71)∧(∀x73:gt(x72, x71)=false∧max(x73)=x71 ⇒ COND2(false, x72, x73)≥ST(s(x72), x73)) ⇒ COND2(false, s(x72), x37)≥ST(s(s(x72)), x37))


    We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint:
    (5)    (COND2(false, 0, x37)≥ST(s(0), x37))


    We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on max(x37)=s(x71) which results in the following new constraint:
    (6)    (if(gt(x75, max(x74)), x75, max(x74))=s(x71)∧gt(x72, x71)=false∧(∀x73:gt(x72, x71)=false∧max(x73)=x71 ⇒ COND2(false, x72, x73)≥ST(s(x72), x73))∧(∀x76,x77,x78:max(x74)=s(x76)∧gt(x77, x76)=false∧(∀x78:gt(x77, x76)=false∧max(x78)=x76 ⇒ COND2(false, x77, x78)≥ST(s(x77), x78)) ⇒ COND2(false, s(x77), x74)≥ST(s(s(x77)), x74)) ⇒ COND2(false, s(x72), cons(x75, x74))≥ST(s(s(x72)), cons(x75, x74)))


    We simplified constraint (6) using rules (IV), (VII) which results in the following new constraint:
    (7)    (gt(x75, x81)=x79∧max(x74)=x80∧if(x79, x75, x80)=s(x71)∧gt(x72, x71)=false∧(∀x73:gt(x72, x71)=false∧max(x73)=x71 ⇒ COND2(false, x72, x73)≥ST(s(x72), x73))∧(∀x76,x77,x78:max(x74)=s(x76)∧gt(x77, x76)=false∧(∀x78:gt(x77, x76)=false∧max(x78)=x76 ⇒ COND2(false, x77, x78)≥ST(s(x77), x78)) ⇒ COND2(false, s(x77), x74)≥ST(s(s(x77)), x74)) ⇒ COND2(false, s(x72), cons(x75, x74))≥ST(s(s(x72)), cons(x75, x74)))


    We simplified constraint (7) using rule (V) (with possible (I) afterwards) using induction on if(x79, x75, x80)=s(x71) which results in the following new constraints:
    (8)    (x83=s(x71)∧gt(x83, x81)=true∧max(x74)=x82∧gt(x72, x71)=false∧(∀x73:gt(x72, x71)=false∧max(x73)=x71 ⇒ COND2(false, x72, x73)≥ST(s(x72), x73))∧(∀x76,x77,x78:max(x74)=s(x76)∧gt(x77, x76)=false∧(∀x78:gt(x77, x76)=false∧max(x78)=x76 ⇒ COND2(false, x77, x78)≥ST(s(x77), x78)) ⇒ COND2(false, s(x77), x74)≥ST(s(s(x77)), x74)) ⇒ COND2(false, s(x72), cons(x83, x74))≥ST(s(s(x72)), cons(x83, x74)))

    (9)    (x84=s(x71)∧gt(x85, x81)=false∧max(x74)=x84∧gt(x72, x71)=false∧(∀x73:gt(x72, x71)=false∧max(x73)=x71 ⇒ COND2(false, x72, x73)≥ST(s(x72), x73))∧(∀x76,x77,x78:max(x74)=s(x76)∧gt(x77, x76)=false∧(∀x78:gt(x77, x76)=false∧max(x78)=x76 ⇒ COND2(false, x77, x78)≥ST(s(x77), x78)) ⇒ COND2(false, s(x77), x74)≥ST(s(s(x77)), x74)) ⇒ COND2(false, s(x72), cons(x85, x74))≥ST(s(s(x72)), cons(x85, x74)))


    We simplified constraint (8) using rules (III), (IV), (VII) which results in the following new constraint:
    (10)    (gt(x72, x71)=false ⇒ COND2(false, s(x72), cons(s(x71), x74))≥ST(s(s(x72)), cons(s(x71), x74)))


    We simplified constraint (9) using rule (III) which results in the following new constraint:
    (11)    (gt(x85, x81)=false∧max(x74)=s(x71)∧gt(x72, x71)=false∧(∀x73:gt(x72, x71)=false∧max(x73)=x71 ⇒ COND2(false, x72, x73)≥ST(s(x72), x73))∧(∀x76,x77,x78:max(x74)=s(x76)∧gt(x77, x76)=false∧(∀x78:gt(x77, x76)=false∧max(x78)=x76 ⇒ COND2(false, x77, x78)≥ST(s(x77), x78)) ⇒ COND2(false, s(x77), x74)≥ST(s(s(x77)), x74)) ⇒ COND2(false, s(x72), cons(x85, x74))≥ST(s(s(x72)), cons(x85, x74)))


    We simplified constraint (10) using rule (V) (with possible (I) afterwards) using induction on gt(x72, x71)=false which results in the following new constraints:
    (12)    (false=false ⇒ COND2(false, s(0), cons(s(x87), x74))≥ST(s(s(0)), cons(s(x87), x74)))

    (13)    (gt(x90, x89)=false∧(∀x91:gt(x90, x89)=false ⇒ COND2(false, s(x90), cons(s(x89), x91))≥ST(s(s(x90)), cons(s(x89), x91))) ⇒ COND2(false, s(s(x90)), cons(s(s(x89)), x74))≥ST(s(s(s(x90))), cons(s(s(x89)), x74)))


    We simplified constraint (12) using rules (I), (II) which results in the following new constraint:
    (14)    (COND2(false, s(0), cons(s(x87), x74))≥ST(s(s(0)), cons(s(x87), x74)))


    We simplified constraint (13) using rule (VI) where we applied the induction hypothesis (∀x91:gt(x90, x89)=false ⇒ COND2(false, s(x90), cons(s(x89), x91))≥ST(s(s(x90)), cons(s(x89), x91))) with σ = [x91 / x74] which results in the following new constraint:
    (15)    (COND2(false, s(x90), cons(s(x89), x74))≥ST(s(s(x90)), cons(s(x89), x74)) ⇒ COND2(false, s(s(x90)), cons(s(s(x89)), x74))≥ST(s(s(s(x90))), cons(s(s(x89)), x74)))


    We simplified constraint (11) using rule (VI) where we applied the induction hypothesis (∀x76,x77,x78:max(x74)=s(x76)∧gt(x77, x76)=false∧(∀x78:gt(x77, x76)=false∧max(x78)=x76 ⇒ COND2(false, x77, x78)≥ST(s(x77), x78)) ⇒ COND2(false, s(x77), x74)≥ST(s(s(x77)), x74)) with σ = [x76 / x71, x77 / x72, x78 / x73] which results in the following new constraint:
    (16)    (gt(x85, x81)=false∧COND2(false, s(x72), x74)≥ST(s(s(x72)), x74) ⇒ COND2(false, s(x72), cons(x85, x74))≥ST(s(s(x72)), cons(x85, x74)))


    We simplified constraint (16) using rule (V) (with possible (I) afterwards) using induction on gt(x85, x81)=false which results in the following new constraints:
    (17)    (false=false∧COND2(false, s(x72), x74)≥ST(s(s(x72)), x74) ⇒ COND2(false, s(x72), cons(0, x74))≥ST(s(s(x72)), cons(0, x74)))

    (18)    (gt(x95, x94)=false∧COND2(false, s(x72), x74)≥ST(s(s(x72)), x74)∧(∀x96,x97:gt(x95, x94)=false∧COND2(false, s(x96), x97)≥ST(s(s(x96)), x97) ⇒ COND2(false, s(x96), cons(x95, x97))≥ST(s(s(x96)), cons(x95, x97))) ⇒ COND2(false, s(x72), cons(s(x95), x74))≥ST(s(s(x72)), cons(s(x95), x74)))


    We simplified constraint (17) using rules (I), (II) which results in the following new constraint:
    (19)    (COND2(false, s(x72), x74)≥ST(s(s(x72)), x74) ⇒ COND2(false, s(x72), cons(0, x74))≥ST(s(s(x72)), cons(0, x74)))


    We simplified constraint (18) using rule (VI) where we applied the induction hypothesis (∀x96,x97:gt(x95, x94)=false∧COND2(false, s(x96), x97)≥ST(s(s(x96)), x97) ⇒ COND2(false, s(x96), cons(x95, x97))≥ST(s(s(x96)), cons(x95, x97))) with σ = [x96 / x72, x97 / x74] which results in the following new constraint:
    (20)    (COND2(false, s(x72), cons(x95, x74))≥ST(s(s(x72)), cons(x95, x74)) ⇒ COND2(false, s(x72), cons(s(x95), x74))≥ST(s(s(x72)), cons(s(x95), x74)))






To summarize, we get the following constraints P≥ for the following pairs.
  • COND1(true, n, l) → ST(s(n), l)
    • (COND1(true, x45, x43)≥ST(s(x45), x43) ⇒ COND1(true, x45, cons(x44, x43))≥ST(s(x45), cons(x44, x43)))
    • (COND1(true, 0, cons(0, x43))≥ST(s(0), cons(0, x43)))
    • (COND1(true, x53, cons(x52, x43))≥ST(s(x53), cons(x52, x43)) ⇒ COND1(true, s(x53), cons(s(x52), x43))≥ST(s(s(x53)), cons(s(x52), x43)))

  • ST(n, l) → COND1(member(n, l), n, l)
    • (ST(s(x10), x11)≥COND1(member(s(x10), x11), s(x10), x11))
    • (ST(s(x18), x19)≥COND1(member(s(x18), x19), s(x18), x19))

  • COND1(false, n, l) → COND2(gt(n, max(l)), n, l)
    • (COND1(false, x55, nil)≥COND2(gt(x55, max(nil)), x55, nil))
    • (COND1(false, s(x63), x56)≥COND2(gt(s(x63), max(x56)), s(x63), x56) ⇒ COND1(false, s(x63), cons(0, x56))≥COND2(gt(s(x63), max(cons(0, x56))), s(x63), cons(0, x56)))
    • (COND1(false, 0, x56)≥COND2(gt(0, max(x56)), 0, x56) ⇒ COND1(false, 0, cons(s(x64), x56))≥COND2(gt(0, max(cons(s(x64), x56))), 0, cons(s(x64), x56)))
    • (COND1(false, s(x66), x56)≥COND2(gt(s(x66), max(x56)), s(x66), x56) ⇒ COND1(false, s(x66), cons(s(x65), x56))≥COND2(gt(s(x66), max(cons(s(x65), x56))), s(x66), cons(s(x65), x56)))

  • COND2(false, n, l) → ST(s(n), l)
    • (COND2(false, 0, x37)≥ST(s(0), x37))
    • (COND2(false, s(0), cons(s(x87), x74))≥ST(s(s(0)), cons(s(x87), x74)))
    • (COND2(false, s(x90), cons(s(x89), x74))≥ST(s(s(x90)), cons(s(x89), x74)) ⇒ COND2(false, s(s(x90)), cons(s(s(x89)), x74))≥ST(s(s(s(x90))), cons(s(s(x89)), x74)))
    • (COND2(false, s(x72), x74)≥ST(s(s(x72)), x74) ⇒ COND2(false, s(x72), cons(0, x74))≥ST(s(s(x72)), cons(0, x74)))
    • (COND2(false, s(x72), cons(x95, x74))≥ST(s(s(x72)), cons(x95, x74)) ⇒ COND2(false, s(x72), cons(s(x95), x74))≥ST(s(s(x72)), cons(s(x95), x74)))




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(COND1(x1, x2, x3)) = 1 - x1 - x2 + x3   
POL(COND2(x1, x2, x3)) = 1 - x1 - x2 + x3   
POL(ST(x1, x2)) = 1 - x1 + x2   
POL(c) = -1   
POL(cons(x1, x2)) = x1 + x2   
POL(equal(x1, x2)) = 0   
POL(false) = 0   
POL(gt(x1, x2)) = 0   
POL(if(x1, x2, x3)) = x3   
POL(max(x1)) = x1   
POL(member(x1, x2)) = 0   
POL(nil) = 1   
POL(or(x1, x2)) = x2   
POL(s(x1)) = 1 + x1   
POL(true) = 1   

The following pairs are in P>:

COND2(false, n, l) → ST(s(n), l)
The following pairs are in Pbound:

COND1(true, n, l) → ST(s(n), l)
COND2(false, n, l) → ST(s(n), l)
The following rules are usable:

false → member(n, nil)
or(equal(n, m), member(n, l)) → member(n, cons(m, l))
false → gt(0, v)
true → gt(s(u), 0)
gt(u, v) → gt(s(u), s(v))
true → or(x, true)
x → or(x, false)

(41) Obligation:

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

COND1(true, n, l) → ST(s(n), l)
ST(n, l) → COND1(member(n, l), n, l)
COND1(false, n, l) → COND2(gt(n, max(l)), n, l)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
if(true, u, v) → u
if(false, u, v) → v
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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

(42) DependencyGraphProof (EQUIVALENT transformation)

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

(43) Obligation:

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

ST(n, l) → COND1(member(n, l), n, l)
COND1(true, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
if(true, u, v) → u
if(false, u, v) → v
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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

(44) 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.

(45) Obligation:

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

ST(n, l) → COND1(member(n, l), n, l)
COND1(true, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

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

(46) 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].

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
max(nil)
max(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)

(47) Obligation:

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

ST(n, l) → COND1(member(n, l), n, l)
COND1(true, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))

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

(48) 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 ST(n, l) → COND1(member(n, l), n, l) the following chains were created:
  • We consider the chain COND1(true, x2, x3) → ST(s(x2), x3), ST(x4, x5) → COND1(member(x4, x5), x4, x5) which results in the following constraint:
    (1)    (ST(s(x2), x3)=ST(x4, x5) ⇒ ST(x4, x5)≥COND1(member(x4, x5), x4, x5))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (ST(s(x2), x3)≥COND1(member(s(x2), x3), s(x2), x3))






For Pair COND1(true, n, l) → ST(s(n), l) the following chains were created:
  • We consider the chain ST(x6, x7) → COND1(member(x6, x7), x6, x7), COND1(true, x8, x9) → ST(s(x8), x9) which results in the following constraint:
    (1)    (COND1(member(x6, x7), x6, x7)=COND1(true, x8, x9) ⇒ COND1(true, x8, x9)≥ST(s(x8), x9))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (member(x6, x7)=true ⇒ COND1(true, x6, x7)≥ST(s(x6), x7))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on member(x6, x7)=true which results in the following new constraint:
    (3)    (or(equal(x15, x14), member(x15, x13))=true∧(member(x15, x13)=true ⇒ COND1(true, x15, x13)≥ST(s(x15), x13)) ⇒ COND1(true, x15, cons(x14, x13))≥ST(s(x15), cons(x14, x13)))


    We simplified constraint (3) using rule (VII) which results in the following new constraint:
    (4)    (equal(x15, x14)=x16∧member(x15, x13)=x17∧or(x16, x17)=true∧(member(x15, x13)=true ⇒ COND1(true, x15, x13)≥ST(s(x15), x13)) ⇒ COND1(true, x15, cons(x14, x13))≥ST(s(x15), cons(x14, x13)))


    We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on or(x16, x17)=true which results in the following new constraints:
    (5)    (true=true∧equal(x15, x14)=x18∧member(x15, x13)=true∧(member(x15, x13)=true ⇒ COND1(true, x15, x13)≥ST(s(x15), x13)) ⇒ COND1(true, x15, cons(x14, x13))≥ST(s(x15), cons(x14, x13)))

    (6)    (x19=true∧equal(x15, x14)=x19∧member(x15, x13)=false∧(member(x15, x13)=true ⇒ COND1(true, x15, x13)≥ST(s(x15), x13)) ⇒ COND1(true, x15, cons(x14, x13))≥ST(s(x15), cons(x14, x13)))


    We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
    (7)    (equal(x15, x14)=x18∧member(x15, x13)=true∧(member(x15, x13)=true ⇒ COND1(true, x15, x13)≥ST(s(x15), x13)) ⇒ COND1(true, x15, cons(x14, x13))≥ST(s(x15), cons(x14, x13)))


    We simplified constraint (6) using rules (III), (IV) which results in the following new constraint:
    (8)    (equal(x15, x14)=true ⇒ COND1(true, x15, cons(x14, x13))≥ST(s(x15), cons(x14, x13)))


    We simplified constraint (7) using rule (VI) where we applied the induction hypothesis (member(x15, x13)=true ⇒ COND1(true, x15, x13)≥ST(s(x15), x13)) with σ = [ ] which results in the following new constraint:
    (9)    (equal(x15, x14)=x18∧COND1(true, x15, x13)≥ST(s(x15), x13) ⇒ COND1(true, x15, cons(x14, x13))≥ST(s(x15), cons(x14, x13)))


    We simplified constraint (9) using rule (IV) which results in the following new constraint:
    (10)    (COND1(true, x15, x13)≥ST(s(x15), x13) ⇒ COND1(true, x15, cons(x14, x13))≥ST(s(x15), cons(x14, x13)))


    We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on equal(x15, x14)=true which results in the following new constraints:
    (11)    (true=true ⇒ COND1(true, 0, cons(0, x13))≥ST(s(0), cons(0, x13)))

    (12)    (equal(x23, x22)=true∧(∀x24:equal(x23, x22)=true ⇒ COND1(true, x23, cons(x22, x24))≥ST(s(x23), cons(x22, x24))) ⇒ COND1(true, s(x23), cons(s(x22), x13))≥ST(s(s(x23)), cons(s(x22), x13)))


    We simplified constraint (11) using rules (I), (II) which results in the following new constraint:
    (13)    (COND1(true, 0, cons(0, x13))≥ST(s(0), cons(0, x13)))


    We simplified constraint (12) using rule (VI) where we applied the induction hypothesis (∀x24:equal(x23, x22)=true ⇒ COND1(true, x23, cons(x22, x24))≥ST(s(x23), cons(x22, x24))) with σ = [x24 / x13] which results in the following new constraint:
    (14)    (COND1(true, x23, cons(x22, x13))≥ST(s(x23), cons(x22, x13)) ⇒ COND1(true, s(x23), cons(s(x22), x13))≥ST(s(s(x23)), cons(s(x22), x13)))






To summarize, we get the following constraints P≥ for the following pairs.
  • ST(n, l) → COND1(member(n, l), n, l)
    • (ST(s(x2), x3)≥COND1(member(s(x2), x3), s(x2), x3))

  • COND1(true, n, l) → ST(s(n), l)
    • (COND1(true, x15, x13)≥ST(s(x15), x13) ⇒ COND1(true, x15, cons(x14, x13))≥ST(s(x15), cons(x14, x13)))
    • (COND1(true, 0, cons(0, x13))≥ST(s(0), cons(0, x13)))
    • (COND1(true, x23, cons(x22, x13))≥ST(s(x23), cons(x22, x13)) ⇒ COND1(true, s(x23), cons(s(x22), x13))≥ST(s(s(x23)), cons(s(x22), x13)))




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) = 0   
POL(COND1(x1, x2, x3)) = -x2 + x3   
POL(ST(x1, x2)) = -x1 + x2   
POL(c) = -1   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(equal(x1, x2)) = 1   
POL(false) = 0   
POL(member(x1, x2)) = 1   
POL(nil) = 0   
POL(or(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following pairs are in P>:

COND1(true, n, l) → ST(s(n), l)
The following pairs are in Pbound:

COND1(true, n, l) → ST(s(n), l)
There are no usable rules

(49) Obligation:

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

ST(n, l) → COND1(member(n, l), n, l)

The TRS R consists of the following rules:

member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
or(x, true) → true
or(x, false) → x

The set Q consists of the following terms:

member(x0, nil)
member(x0, cons(x1, x2))
or(x0, true)
or(x0, false)
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))

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

(50) DependencyGraphProof (EQUIVALENT transformation)

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

(51) TRUE