YES Termination w.r.t. Q proof of Secret_07_TRS_aprove07.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 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 NonInfProof (⇔, 0 ms)
        ↳20 AND
            ↳21 QDP
                ↳22 DependencyGraphProof (⇔, 0 ms)
                ↳23 TRUE
            ↳24 QDP
                ↳25 DependencyGraphProof (⇔, 0 ms)
                ↳26 TRUE
    ↳27 QDP
        ↳28 UsableRulesProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QReductionProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QDPOrderProof (⇔, 22 ms)
        ↳33 QDP
        ↳34 DependencyGraphProof (⇔, 0 ms)
        ↳35 TRUE

(0) Obligation:

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

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))

The TRS R 2 is

a → b
a → c

The signature Sigma is {a, b, c}

(2) Obligation:

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

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

PLUS(x, y) → PLUSITER(x, y, 0)
PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)
PLUSITER(x, y, z) → LE(x, z)
IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))
LE(s(x), s(y)) → LE(x, y)
SUM(xs) → SUMITER(xs, 0)
SUMITER(xs, x) → IFSUM(isempty(xs), xs, x, plus(x, head(xs)))
SUMITER(xs, x) → ISEMPTY(xs)
SUMITER(xs, x) → PLUS(x, head(xs))
SUMITER(xs, x) → HEAD(xs)
IFSUM(false, xs, x, y) → SUMITER(tail(xs), y)
IFSUM(false, xs, x, y) → TAIL(xs)

The TRS R consists of the following rules:

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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:

LE(s(x), s(y)) → LE(x, y)

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

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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

(10) QReductionProof (EQUIVALENT transformation)

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

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

(11) Obligation:

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

LE(s(x), s(y)) → LE(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:

  • LE(s(x), s(y)) → LE(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:

IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))
PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)

The TRS R consists of the following rules:

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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:

IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))
PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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

(17) QReductionProof (EQUIVALENT transformation)

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

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

(18) Obligation:

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

IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))
PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))

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

(19) 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 IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z)) the following chains were created:
  • We consider the chain PLUSITER(x3, x4, x5) → IFPLUS(le(x3, x5), x3, x4, x5), IFPLUS(false, x6, x7, x8) → PLUSITER(x6, s(x7), s(x8)) which results in the following constraint:
    (1)    (IFPLUS(le(x3, x5), x3, x4, x5)=IFPLUS(false, x6, x7, x8) ⇒ IFPLUS(false, x6, x7, x8)≥PLUSITER(x6, s(x7), s(x8)))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (le(x3, x5)=false ⇒ IFPLUS(false, x3, x4, x5)≥PLUSITER(x3, s(x4), s(x5)))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x3, x5)=false which results in the following new constraints:
    (3)    (false=false ⇒ IFPLUS(false, s(x18), x4, 0)≥PLUSITER(s(x18), s(x4), s(0)))

    (4)    (le(x21, x20)=false∧(∀x22:le(x21, x20)=false ⇒ IFPLUS(false, x21, x22, x20)≥PLUSITER(x21, s(x22), s(x20))) ⇒ IFPLUS(false, s(x21), x4, s(x20))≥PLUSITER(s(x21), s(x4), s(s(x20))))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (IFPLUS(false, s(x18), x4, 0)≥PLUSITER(s(x18), s(x4), s(0)))


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x22:le(x21, x20)=false ⇒ IFPLUS(false, x21, x22, x20)≥PLUSITER(x21, s(x22), s(x20))) with σ = [x22 / x4] which results in the following new constraint:
    (6)    (IFPLUS(false, x21, x4, x20)≥PLUSITER(x21, s(x4), s(x20)) ⇒ IFPLUS(false, s(x21), x4, s(x20))≥PLUSITER(s(x21), s(x4), s(s(x20))))






For Pair PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z) the following chains were created:
  • We consider the chain IFPLUS(false, x9, x10, x11) → PLUSITER(x9, s(x10), s(x11)), PLUSITER(x12, x13, x14) → IFPLUS(le(x12, x14), x12, x13, x14) which results in the following constraint:
    (1)    (PLUSITER(x9, s(x10), s(x11))=PLUSITER(x12, x13, x14) ⇒ PLUSITER(x12, x13, x14)≥IFPLUS(le(x12, x14), x12, x13, x14))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (PLUSITER(x9, s(x10), s(x11))≥IFPLUS(le(x9, s(x11)), x9, s(x10), s(x11)))






To summarize, we get the following constraints P≥ for the following pairs.
  • IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))
    • (IFPLUS(false, s(x18), x4, 0)≥PLUSITER(s(x18), s(x4), s(0)))
    • (IFPLUS(false, x21, x4, x20)≥PLUSITER(x21, s(x4), s(x20)) ⇒ IFPLUS(false, s(x21), x4, s(x20))≥PLUSITER(s(x21), s(x4), s(s(x20))))

  • PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)
    • (PLUSITER(x9, s(x10), s(x11))≥IFPLUS(le(x9, s(x11)), x9, s(x10), s(x11)))




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(IFPLUS(x1, x2, x3, x4)) = x2 - x4   
POL(PLUSITER(x1, x2, x3)) = 1 + x1 - x3   
POL(c) = -1   
POL(false) = 1   
POL(le(x1, x2)) = 1 + x2   
POL(s(x1)) = 1 + x1   
POL(true) = 1   

The following pairs are in P>:

PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)
The following pairs are in Pbound:

IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))
There are no usable rules

(20) Complex Obligation (AND)

(21) Obligation:

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

IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))

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

(22) DependencyGraphProof (EQUIVALENT transformation)

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

(23) TRUE

(24) Obligation:

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

PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))

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

(25) DependencyGraphProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

SUMITER(xs, x) → IFSUM(isempty(xs), xs, x, plus(x, head(xs)))
IFSUM(false, xs, x, y) → SUMITER(tail(xs), y)

The TRS R consists of the following rules:

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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

(28) 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.

(29) Obligation:

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

SUMITER(xs, x) → IFSUM(isempty(xs), xs, x, plus(x, head(xs)))
IFSUM(false, xs, x, y) → SUMITER(tail(xs), y)

The TRS R consists of the following rules:

tail(nil) → nil
tail(cons(x, xs)) → xs
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
plus(x, y) → plusIter(x, y, 0)
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
ifPlus(true, x, y, z) → y

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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

(30) 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].

sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
a

(31) Obligation:

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

SUMITER(xs, x) → IFSUM(isempty(xs), xs, x, plus(x, head(xs)))
IFSUM(false, xs, x, y) → SUMITER(tail(xs), y)

The TRS R consists of the following rules:

tail(nil) → nil
tail(cons(x, xs)) → xs
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
plus(x, y) → plusIter(x, y, 0)
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
ifPlus(true, x, y, z) → y

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

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

(32) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


IFSUM(false, xs, x, y) → SUMITER(tail(xs), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(0) = 0   
POL(IFSUM(x1, x2, x3, x4)) = [1/4] + x1 + [1/4]x2   
POL(SUMITER(x1, x2)) = [1/2] + [1/2]x1   
POL(cons(x1, x2)) = [4] + [4]x1 + [2]x2   
POL(error) = 0   
POL(false) = [1]   
POL(head(x1)) = 0   
POL(ifPlus(x1, x2, x3, x4)) = [1/2] + x2 + [4]x3 + x4   
POL(isempty(x1)) = [1/4] + [1/4]x1   
POL(le(x1, x2)) = [1] + [1/4]x2   
POL(nil) = [1/2]   
POL(plus(x1, x2)) = [4] + x1 + [4]x2   
POL(plusIter(x1, x2, x3)) = [1/2] + x1 + [4]x2 + x3   
POL(s(x1)) = x1   
POL(tail(x1)) = [1] + [1/2]x1   
POL(true) = 0   
The value of delta used in the strict ordering is 1/4.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

isempty(nil) → true
isempty(cons(x, xs)) → false
tail(nil) → nil
tail(cons(x, xs)) → xs

(33) Obligation:

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

SUMITER(xs, x) → IFSUM(isempty(xs), xs, x, plus(x, head(xs)))

The TRS R consists of the following rules:

tail(nil) → nil
tail(cons(x, xs)) → xs
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
plus(x, y) → plusIter(x, y, 0)
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
ifPlus(true, x, y, z) → y

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

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

(34) DependencyGraphProof (EQUIVALENT transformation)

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

(35) TRUE