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

(0) Obligation:

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

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(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:

LT(s(x), s(y)) → LT(x, y)
FIBO(0) → FIB(0)
FIBO(s(0)) → FIB(s(0))
FIBO(s(s(x))) → SUM(fibo(s(x)), fibo(x))
FIBO(s(s(x))) → FIBO(s(x))
FIBO(s(s(x))) → FIBO(x)
FIB(s(s(x))) → IF(true, 0, s(s(x)), 0, 0)
IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
IF(true, c, s(s(x)), a, b) → LT(s(c), s(s(x)))
IF(false, c, s(s(x)), a, b) → SUM(fibo(a), fibo(b))
IF(false, c, s(s(x)), a, b) → FIBO(a)
IF(false, c, s(s(x)), a, b) → FIBO(b)
SUM(x, s(y)) → SUM(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(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 4 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

SUM(x, s(y)) → SUM(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(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:

SUM(x, s(y)) → SUM(x, y)

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

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(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].

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

(11) Obligation:

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

SUM(x, s(y)) → SUM(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:

• SUM(x, s(y)) → SUM(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:

FIBO(s(s(x))) → FIBO(s(x))
FIBO(s(s(x))) → FIBO(x)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(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:

FIBO(s(s(x))) → FIBO(s(x))
FIBO(s(s(x))) → FIBO(x)

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

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(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].

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

(18) Obligation:

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

FIBO(s(s(x))) → FIBO(s(x))
FIBO(s(s(x))) → FIBO(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:

• FIBO(s(s(x))) → FIBO(s(x))
  The graph contains the following edges 1 > 1

• FIBO(s(s(x))) → FIBO(x)
  The graph contains the following edges 1 > 1

(20) TRUE

(21) Obligation:

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

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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:

LT(s(x), s(y)) → LT(x, y)

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

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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].

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

(25) Obligation:

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

LT(s(x), s(y)) → LT(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:

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

(27) TRUE

(28) Obligation:

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

IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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:

IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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].

fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

(32) Obligation:

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

IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

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

(33) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c) we obtained the following new rules [LPAR04]:

IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(s(s(z0)), s(s(z1))), s(s(z0)), s(s(z1)), z0, s(z0))

(34) Obligation:

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

IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(s(s(z0)), s(s(z1))), s(s(z0)), s(s(z1)), z0, s(z0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

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

(35) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c) at position [0] we obtained the following new rules [LPAR04]:

IF(true, c, s(s(x)), a, b) → IF(lt(c, s(x)), s(c), s(s(x)), b, c)

(36) Obligation:

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

IF(true, c, s(s(x)), a, b) → IF(lt(c, s(x)), s(c), s(s(x)), b, c)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

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

(37) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(true, c, s(s(x)), a, b) → IF(lt(c, s(x)), s(c), s(s(x)), b, c) we obtained the following new rules [LPAR04]:

IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(s(z0), s(z1)), s(s(z0)), s(s(z1)), z0, s(z0))

(38) Obligation:

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

IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(s(z0), s(z1)), s(s(z0)), s(s(z1)), z0, s(z0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

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

(39) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(s(z0), s(z1)), s(s(z0)), s(s(z1)), z0, s(z0)) at position [0] we obtained the following new rules [LPAR04]:

IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(z0, z1), s(s(z0)), s(s(z1)), z0, s(z0))

(40) Obligation:

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

IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(z0, z1), s(s(z0)), s(s(z1)), z0, s(z0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

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

(41) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(z0, z1), s(s(z0)), s(s(z1)), z0, s(z0)) we obtained the following new rules [LPAR04]:

IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0)))

(42) Obligation:

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

IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

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

(43) 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(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0))) the following chains were created:

• We consider the chain IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0))), IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0))) which results in the following constraint:
  

  (1)    (IF(lt(s(x0), x1), s(s(s(x0))), s(s(x1)), s(x0), s(s(x0)))=IF(true, s(s(x2)), s(s(x3)), x2, s(x2)) ⇒ IF(true, s(s(x2)), s(s(x3)), x2, s(x2))≥IF(lt(s(x2), x3), s(s(s(x2))), s(s(x3)), s(x2), s(s(x2))))

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

  (2)    (s(x0)=x4∧lt(x4, x1)=true ⇒ IF(true, s(s(s(x0))), s(s(x1)), s(x0), s(s(x0)))≥IF(lt(s(s(x0)), x1), s(s(s(s(x0)))), s(s(x1)), s(s(x0)), s(s(s(x0)))))

  
  
  We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on lt(x4, x1)=true which results in the following new constraints:
  

  (3)    (lt(x6, x5)=true∧s(x0)=s(x6)∧(∀x7:lt(x6, x5)=true∧s(x7)=x6 ⇒ IF(true, s(s(s(x7))), s(s(x5)), s(x7), s(s(x7)))≥IF(lt(s(s(x7)), x5), s(s(s(s(x7)))), s(s(x5)), s(s(x7)), s(s(s(x7))))) ⇒ IF(true, s(s(s(x0))), s(s(s(x5))), s(x0), s(s(x0)))≥IF(lt(s(s(x0)), s(x5)), s(s(s(s(x0)))), s(s(s(x5))), s(s(x0)), s(s(s(x0)))))

  
  

  (4)    (true=true∧s(x0)=0 ⇒ IF(true, s(s(s(x0))), s(s(s(x8))), s(x0), s(s(x0)))≥IF(lt(s(s(x0)), s(x8)), s(s(s(s(x0)))), s(s(s(x8))), s(s(x0)), s(s(s(x0)))))

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

  (5)    (lt(x6, x5)=true ⇒ IF(true, s(s(s(x6))), s(s(s(x5))), s(x6), s(s(x6)))≥IF(lt(s(s(x6)), s(x5)), s(s(s(s(x6)))), s(s(s(x5))), s(s(x6)), s(s(s(x6)))))

  
  
  We solved constraint (4) using rules (I), (II).We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on lt(x6, x5)=true which results in the following new constraints:
  

  (6)    (lt(x11, x10)=true∧(lt(x11, x10)=true ⇒ IF(true, s(s(s(x11))), s(s(s(x10))), s(x11), s(s(x11)))≥IF(lt(s(s(x11)), s(x10)), s(s(s(s(x11)))), s(s(s(x10))), s(s(x11)), s(s(s(x11))))) ⇒ IF(true, s(s(s(s(x11)))), s(s(s(s(x10)))), s(s(x11)), s(s(s(x11))))≥IF(lt(s(s(s(x11))), s(s(x10))), s(s(s(s(s(x11))))), s(s(s(s(x10)))), s(s(s(x11))), s(s(s(s(x11))))))

  
  

  (7)    (true=true ⇒ IF(true, s(s(s(0))), s(s(s(s(x12)))), s(0), s(s(0)))≥IF(lt(s(s(0)), s(s(x12))), s(s(s(s(0)))), s(s(s(s(x12)))), s(s(0)), s(s(s(0)))))

  
  
  We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (lt(x11, x10)=true ⇒ IF(true, s(s(s(x11))), s(s(s(x10))), s(x11), s(s(x11)))≥IF(lt(s(s(x11)), s(x10)), s(s(s(s(x11)))), s(s(s(x10))), s(s(x11)), s(s(s(x11))))) with σ = [ ] which results in the following new constraint:
  

  (8)    (IF(true, s(s(s(x11))), s(s(s(x10))), s(x11), s(s(x11)))≥IF(lt(s(s(x11)), s(x10)), s(s(s(s(x11)))), s(s(s(x10))), s(s(x11)), s(s(s(x11)))) ⇒ IF(true, s(s(s(s(x11)))), s(s(s(s(x10)))), s(s(x11)), s(s(s(x11))))≥IF(lt(s(s(s(x11))), s(s(x10))), s(s(s(s(s(x11))))), s(s(s(s(x10)))), s(s(s(x11))), s(s(s(s(x11))))))

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

  (9)    (IF(true, s(s(s(0))), s(s(s(s(x12)))), s(0), s(s(0)))≥IF(lt(s(s(0)), s(s(x12))), s(s(s(s(0)))), s(s(s(s(x12)))), s(s(0)), s(s(s(0)))))

  
  
  

To summarize, we get the following constraints P_≥ for the following pairs.

• IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0)))
  
  • (IF(true, s(s(s(x11))), s(s(s(x10))), s(x11), s(s(x11)))≥IF(lt(s(s(x11)), s(x10)), s(s(s(s(x11)))), s(s(s(x10))), s(s(x11)), s(s(s(x11)))) ⇒ IF(true, s(s(s(s(x11)))), s(s(s(s(x10)))), s(s(x11)), s(s(s(x11))))≥IF(lt(s(s(s(x11))), s(s(x10))), s(s(s(s(s(x11))))), s(s(s(s(x10)))), s(s(s(x11))), s(s(s(s(x11))))))
    
  • (IF(true, s(s(s(0))), s(s(s(s(x12)))), s(0), s(s(0)))≥IF(lt(s(s(0)), s(s(x12))), s(s(s(s(0)))), s(s(s(s(x12)))), s(s(0)), s(s(s(0)))))
    
  
  

The constraints for P_> respective P_bound 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 P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(IF(x₁, x₂, x₃, x₄, x₅)) = -1 - x₁ - x₂ + x₃ - x₄ + x₅   
POL(c) = -1   
POL(false) = 0   
POL(lt(x₁, x₂)) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following pairs are in P_>:

IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0)))

The following pairs are in P_bound:

IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0)))

The following rules are usable:

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

(44) Obligation:

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

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

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

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE