YES Termination w.r.t. Q proof of AProVE_07_thiemann03.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 5 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 TransformationProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 DependencyGraphProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 UsableRulesProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QReductionProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 TransformationProof (⇔, 0 ms)
        ↳35 QDP
        ↳36 UsableRulesProof (⇔, 0 ms)
        ↳37 QDP
        ↳38 QReductionProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 TransformationProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 DependencyGraphProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 UsableRulesProof (⇔, 0 ms)
        ↳45 QDP
        ↳46 QReductionProof (⇔, 0 ms)
        ↳47 QDP
        ↳48 QDPOrderProof (⇔, 9 ms)
        ↳49 QDP
        ↳50 DependencyGraphProof (⇔, 0 ms)
        ↳51 TRUE

(0) Obligation:

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

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
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(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), 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:

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
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(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

The set Q consists of the following terms:

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(x0)
shuff(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

(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:

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(x) → SHUFF(x, nil)
SHUFF(x, y) → IF(null(x), x, y, app(y, add(head(x), nil)))
SHUFF(x, y) → NULL(x)
SHUFF(x, y) → APP(y, add(head(x), nil))
SHUFF(x, y) → HEAD(x)
IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)
IF(false, x, y, z) → REVERSE(tail(x))
IF(false, x, y, z) → TAIL(x)

The TRS R consists of the following rules:

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
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(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

The set Q consists of the following terms:

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(x0)
shuff(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 7 less nodes.

(6) Complex Obligation (AND)

(7) 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:

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
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(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

The set Q consists of the following terms:

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(x0)
shuff(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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:

APP(add(n, x), y) → APP(x, y)

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

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(x0)
shuff(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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].

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(x0)
shuff(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

(11) 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.

(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:

  • APP(add(n, x), y) → APP(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:

REVERSE(add(n, x)) → REVERSE(x)

The TRS R consists of the following rules:

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
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(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

The set Q consists of the following terms:

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(x0)
shuff(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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:

REVERSE(add(n, x)) → REVERSE(x)

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

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(x0)
shuff(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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].

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(x0)
shuff(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

(18) 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.

(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:

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

(20) YES

(21) Obligation:

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

IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)
SHUFF(x, y) → IF(null(x), x, y, app(y, add(head(x), nil)))

The TRS R consists of the following rules:

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
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(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

The set Q consists of the following terms:

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(x0)
shuff(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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:

IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)
SHUFF(x, y) → IF(null(x), x, y, app(y, add(head(x), nil)))

The TRS R consists of the following rules:

null(nil) → true
null(add(n, x)) → false
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
tail(add(n, x)) → x
tail(nil) → nil
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))

The set Q consists of the following terms:

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(x0)
shuff(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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].

shuffle(x0)
shuff(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

(25) Obligation:

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

IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)
SHUFF(x, y) → IF(null(x), x, y, app(y, add(head(x), nil)))

The TRS R consists of the following rules:

null(nil) → true
null(add(n, x)) → false
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
tail(add(n, x)) → x
tail(nil) → nil
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))

The set Q consists of the following terms:

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))

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

(26) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule SHUFF(x, y) → IF(null(x), x, y, app(y, add(head(x), nil))) at position [0] we obtained the following new rules [LPAR04]:

SHUFF(nil, y1) → IF(true, nil, y1, app(y1, add(head(nil), nil))) → SHUFF(nil, y1) → IF(true, nil, y1, app(y1, add(head(nil), nil)))
SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(head(add(x0, x1)), nil))) → SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(head(add(x0, x1)), nil)))

(27) Obligation:

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

IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)
SHUFF(nil, y1) → IF(true, nil, y1, app(y1, add(head(nil), nil)))
SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(head(add(x0, x1)), nil)))

The TRS R consists of the following rules:

null(nil) → true
null(add(n, x)) → false
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
tail(add(n, x)) → x
tail(nil) → nil
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))

The set Q consists of the following terms:

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))

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

(28) DependencyGraphProof (EQUIVALENT transformation)

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

(29) Obligation:

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

SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(head(add(x0, x1)), nil)))
IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)

The TRS R consists of the following rules:

null(nil) → true
null(add(n, x)) → false
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
tail(add(n, x)) → x
tail(nil) → nil
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))

The set Q consists of the following terms:

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))

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

(30) 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.

(31) Obligation:

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

SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(head(add(x0, x1)), nil)))
IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
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))
head(add(n, x)) → n

The set Q consists of the following terms:

null(nil)
null(add(x0, x1))
tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))

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

(32) 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].

null(nil)
null(add(x0, x1))

(33) Obligation:

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

SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(head(add(x0, x1)), nil)))
IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
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))
head(add(n, x)) → n

The set Q consists of the following terms:

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

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

(34) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(head(add(x0, x1)), nil))) at position [3,1,0] we obtained the following new rules [LPAR04]:

SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(x0, nil))) → SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(x0, nil)))

(35) Obligation:

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

IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)
SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(x0, nil)))

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
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))
head(add(n, x)) → n

The set Q consists of the following terms:

tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(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:

IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)
SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(x0, nil)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

tail(add(x0, x1))
tail(nil)
head(add(x0, x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(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].

head(add(x0, x1))

(39) Obligation:

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

IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)
SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(x0, nil)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(40) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF(false, x, y, z) → SHUFF(reverse(tail(x)), z) at position [0] we obtained the following new rules [LPAR04]:

IF(false, add(x0, x1), y1, y2) → SHUFF(reverse(x1), y2) → IF(false, add(x0, x1), y1, y2) → SHUFF(reverse(x1), y2)
IF(false, nil, y1, y2) → SHUFF(reverse(nil), y2) → IF(false, nil, y1, y2) → SHUFF(reverse(nil), y2)

(41) Obligation:

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

SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(x0, nil)))
IF(false, add(x0, x1), y1, y2) → SHUFF(reverse(x1), y2)
IF(false, nil, y1, y2) → SHUFF(reverse(nil), y2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

tail(add(x0, x1))
tail(nil)
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) 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:

IF(false, add(x0, x1), y1, y2) → SHUFF(reverse(x1), y2)
SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(x0, nil)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

tail(add(x0, x1))
tail(nil)
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) 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:

IF(false, add(x0, x1), y1, y2) → SHUFF(reverse(x1), y2)
SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(x0, nil)))

The TRS R consists of the following rules:

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))

The set Q consists of the following terms:

tail(add(x0, x1))
tail(nil)
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(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].

tail(add(x0, x1))
tail(nil)

(47) Obligation:

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

IF(false, add(x0, x1), y1, y2) → SHUFF(reverse(x1), y2)
SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(x0, nil)))

The TRS R consists of the following rules:

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))

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.

(48) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


IF(false, add(x0, x1), y1, y2) → SHUFF(reverse(x1), y2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( SHUFF(x1, x2) ) = 2x1 + 2

POL( reverse(x1) ) = x1

POL( nil ) = 0

POL( add(x1, x2) ) = x2 + 1

POL( app(x1, x2) ) = x1 + x2

POL( IF(x1, ..., x4) ) = 2x1 + 2x2

POL( false ) = 1


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

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))

(49) Obligation:

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

SHUFF(add(x0, x1), y1) → IF(false, add(x0, x1), y1, app(y1, add(x0, nil)))

The TRS R consists of the following rules:

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))

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.

(50) DependencyGraphProof (EQUIVALENT transformation)

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

(51) TRUE