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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
    ↳19 UsableRulesReductionPairsProof (⇔)
    ↳20 QDP
    ↳21 PisEmptyProof (⇔)
    ↳22 TRUE
  ↳23 QDP
    ↳24 UsableRulesProof (⇔)
    ↳25 QDP
    ↳26 QReductionProof (⇔)
    ↳27 QDP
      ↳28 Instantiation (⇔)
        ↳29 QDP
      ↳30 NonInfProof (⇔)
        ↳31 QDP
        ↳32 DependencyGraphProof (⇔)
        ↳33 TRUE

(0) Obligation:

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

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

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:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(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:

TIMES(x, y) → SUM(generate(x, y))
TIMES(x, y) → GENERATE(x, y)
GENERATE(x, y) → GEN(x, y, 0)
GEN(x, y, z) → IF(ge(z, x), x, y, z)
GEN(x, y, z) → GE(z, x)
IF(false, x, y, z) → GEN(x, y, s(z))
SUM(cons(0, xs)) → SUM(xs)
SUM(cons(s(x), xs)) → SUM(cons(x, xs))
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(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 3 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(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:

GE(s(x), s(y)) → GE(x, y)

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

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(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].

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(11) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

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

(13) TRUE

(14) Obligation:

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

SUM(cons(s(x), xs)) → SUM(cons(x, xs))
SUM(cons(0, xs)) → SUM(xs)

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(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:

SUM(cons(s(x), xs)) → SUM(cons(x, xs))
SUM(cons(0, xs)) → SUM(xs)

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

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(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].

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(18) Obligation:

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

SUM(cons(s(x), xs)) → SUM(cons(x, xs))
SUM(cons(0, xs)) → SUM(xs)

R is empty.
Q is empty.
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:

SUM(cons(s(x), xs)) → SUM(cons(x, xs))
SUM(cons(0, xs)) → SUM(xs)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(SUM(x1)) = 2·x1   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(s(x1)) = 2·x1   

(20) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
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) TRUE

(23) Obligation:

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

IF(false, x, y, z) → GEN(x, y, s(z))
GEN(x, y, z) → IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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:

IF(false, x, y, z) → GEN(x, y, s(z))
GEN(x, y, z) → IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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].

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(x0), x1))

(27) Obligation:

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

IF(false, x, y, z) → GEN(x, y, s(z))
GEN(x, y, z) → IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(28) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule GEN(x, y, z) → IF(ge(z, x), x, y, z) we obtained the following new rules [LPAR04]:

GEN(z0, z1, s(z2)) → IF(ge(s(z2), z0), z0, z1, s(z2))

(29) Obligation:

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

IF(false, x, y, z) → GEN(x, y, s(z))
GEN(z0, z1, s(z2)) → IF(ge(s(z2), z0), z0, z1, s(z2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(30) 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 IF(false, x, y, z) → GEN(x, y, s(z)) the following chains were created:
  • We consider the chain GEN(x, y, z) → IF(ge(z, x), x, y, z), IF(false, x, y, z) → GEN(x, y, s(z)) which results in the following constraint:

    (1)    (IF(ge(x5, x3), x3, x4, x5)=IF(false, x6, x7, x8) ⇒ IF(false, x6, x7, x8)≥GEN(x6, x7, s(x8)))



    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:

    (2)    (ge(x5, x3)=false ⇒ IF(false, x3, x4, x5)≥GEN(x3, x4, s(x5)))



    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x5, x3)=false which results in the following new constraints:

    (3)    (false=false ⇒ IF(false, s(x19), x4, 0)≥GEN(s(x19), x4, s(0)))


    (4)    (ge(x21, x20)=false∧(∀x22:ge(x21, x20)=false ⇒ IF(false, x20, x22, x21)≥GEN(x20, x22, s(x21))) ⇒ IF(false, s(x20), x4, s(x21))≥GEN(s(x20), x4, s(s(x21))))



    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:

    (5)    (IF(false, s(x19), x4, 0)≥GEN(s(x19), x4, s(0)))



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

    (6)    (IF(false, x20, x4, x21)≥GEN(x20, x4, s(x21)) ⇒ IF(false, s(x20), x4, s(x21))≥GEN(s(x20), x4, s(s(x21))))







For Pair GEN(x, y, z) → IF(ge(z, x), x, y, z) the following chains were created:
  • We consider the chain IF(false, x, y, z) → GEN(x, y, s(z)), GEN(x, y, z) → IF(ge(z, x), x, y, z) which results in the following constraint:

    (7)    (GEN(x9, x10, s(x11))=GEN(x12, x13, x14) ⇒ GEN(x12, x13, x14)≥IF(ge(x14, x12), x12, x13, x14))



    We simplified constraint (7) using rules (I), (II), (III) which results in the following new constraint:

    (8)    (GEN(x9, x10, s(x11))≥IF(ge(s(x11), x9), x9, x10, s(x11)))







To summarize, we get the following constraints P≥ for the following pairs.
  • IF(false, x, y, z) → GEN(x, y, s(z))
    • (IF(false, s(x19), x4, 0)≥GEN(s(x19), x4, s(0)))
    • (IF(false, x20, x4, x21)≥GEN(x20, x4, s(x21)) ⇒ IF(false, s(x20), x4, s(x21))≥GEN(s(x20), x4, s(s(x21))))

  • GEN(x, y, z) → IF(ge(z, x), x, y, z)
    • (GEN(x9, x10, s(x11))≥IF(ge(s(x11), x9), x9, 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) = 1   
POL(GEN(x1, x2, x3)) = -1 + x1 - x3   
POL(IF(x1, x2, x3, x4)) = -1 + x2 - x4   
POL(c) = -2   
POL(false) = 1   
POL(ge(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following pairs are in P>:

IF(false, x, y, z) → GEN(x, y, s(z))
The following pairs are in Pbound:

IF(false, x, y, z) → GEN(x, y, s(z))
There are no usable rules

(31) Obligation:

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

GEN(x, y, z) → IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(32) DependencyGraphProof (EQUIVALENT transformation)

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

(33) TRUE