YES Termination w.r.t. Q proof of AG01_3.53.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 4 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 UsableRulesReductionPairsProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 PisEmptyProof (⇔, 0 ms)
        ↳22 YES
    ↳23 QDP
        ↳24 UsableRulesProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QReductionProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QDPSizeChangeProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QReductionProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QDPSizeChangeProof (⇔, 0 ms)
        ↳36 YES
    ↳37 QDP
        ↳38 UsableRulesProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 QReductionProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 MRRProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 PisEmptyProof (⇔, 0 ms)
        ↳45 YES
    ↳46 QDP
        ↳47 UsableRulesProof (⇔, 0 ms)
        ↳48 QDP
        ↳49 QReductionProof (⇔, 0 ms)
        ↳50 QDP
        ↳51 QDPSizeChangeProof (⇔, 0 ms)
        ↳52 YES
    ↳53 QDP
        ↳54 UsableRulesProof (⇔, 0 ms)
        ↳55 QDP
        ↳56 QReductionProof (⇔, 0 ms)
        ↳57 QDP
        ↳58 QDPOrderProof (⇔, 0 ms)
        ↳59 QDP
        ↳60 PisEmptyProof (⇔, 0 ms)
        ↳61 YES

(0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))

The TRS R 2 is

less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

The signature Sigma is {less_leaves, false, true}

(2) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(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:

MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) → MINUS(x, y)
APP(add(n, x), y) → APP(x, y)
REVERSE(add(n, x)) → APP(reverse(x), add(n, nil))
REVERSE(add(n, x)) → REVERSE(x)
SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))
SHUFFLE(add(n, x)) → REVERSE(x)
CONCAT(cons(u, v), y) → CONCAT(v, y)
LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
LESS_LEAVES(cons(u, v), cons(w, z)) → CONCAT(u, v)
LESS_LEAVES(cons(u, v), cons(w, z)) → CONCAT(w, z)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(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 7 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

CONCAT(cons(u, v), y) → CONCAT(v, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(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:

CONCAT(cons(u, v), y) → CONCAT(v, y)

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

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(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].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

(11) Obligation:

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

CONCAT(cons(u, v), y) → CONCAT(v, 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:

  • CONCAT(cons(u, v), y) → CONCAT(v, 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:

LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(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:

LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))

The TRS R consists of the following rules:

concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(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].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

(18) Obligation:

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

LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))

The TRS R consists of the following rules:

concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))

The set Q consists of the following terms:

concat(leaf, x0)
concat(cons(x0, x1), x2)

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

(19) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
The following rules are removed from R:

concat(leaf, y) → y
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(LESS_LEAVES(x1, x2)) = x1 + x2   
POL(concat(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = 2 + 2·x1 + x2   
POL(leaf) = 0   

(20) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

concat(cons(u, v), y) → cons(u, concat(v, y))

The set Q consists of the following terms:

concat(leaf, x0)
concat(cons(x0, x1), x2)

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

(21) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(22) YES

(23) Obligation:

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

APP(add(n, x), y) → APP(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(24) 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.

(25) Obligation:

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

APP(add(n, x), y) → APP(x, y)

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

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(26) 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].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

(27) Obligation:

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

APP(add(n, x), y) → APP(x, y)

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

(28) 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:

  • APP(add(n, x), y) → APP(x, y)
    The graph contains the following edges 1 > 1, 2 >= 2

(29) YES

(30) Obligation:

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

REVERSE(add(n, x)) → REVERSE(x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(31) 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.

(32) Obligation:

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

REVERSE(add(n, x)) → REVERSE(x)

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

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(33) 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].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

(34) Obligation:

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

REVERSE(add(n, x)) → REVERSE(x)

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

(35) 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:

  • REVERSE(add(n, x)) → REVERSE(x)
    The graph contains the following edges 1 > 1

(36) YES

(37) Obligation:

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

SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(38) 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.

(39) Obligation:

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

SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))

The TRS R consists of the following rules:

reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(40) 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].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

(41) Obligation:

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

SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))

The TRS R consists of the following rules:

reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))

The set Q consists of the following terms:

app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))

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

(42) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))


Used ordering: Polynomial interpretation [POLO]:

POL(SHUFFLE(x1)) = 2·x1   
POL(add(x1, x2)) = 1 + x1 + x2   
POL(app(x1, x2)) = x1 + x2   
POL(nil) = 0   
POL(reverse(x1)) = x1   

(43) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))

The set Q consists of the following terms:

app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))

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

(44) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(45) YES

(46) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(47) 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.

(48) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

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

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(49) 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].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

(50) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

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

(51) 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:

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

(52) YES

(53) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(54) 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.

(55) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(56) 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].

quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

(57) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))

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

(58) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( QUOT(x1, x2) ) = max{0, 2x1 - 2}

POL( minus(x1, x2) ) = x1

POL( 0 ) = 0

POL( s(x1) ) = 2x1 + 2


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

(59) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))

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

(60) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(61) YES