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:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

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:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

(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:

1024¹ → 1024_1¹(0)
1024_1¹(x) → IF(lt(x, 10), x)
1024_1¹(x) → LT(x, 10)
1024_1¹(x) → 10¹
IF(true, x) → DOUBLE(1024_1(s(x)))
IF(true, x) → 1024_1¹(s(x))
LT(s(x), s(y)) → LT(x, y)
DOUBLE(s(x)) → DOUBLE(x)
10¹ → DOUBLE(s(double(s(s(0)))))
10¹ → DOUBLE(s(s(0)))

The TRS R consists of the following rules:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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:

DOUBLE(s(x)) → DOUBLE(x)

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

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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].

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

(11) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

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:

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

(13) TRUE

(14) 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:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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:

LT(s(x), s(y)) → LT(x, y)

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

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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].

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

(18) 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.

(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:

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

(20) TRUE

(21) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, 10), x)

The TRS R consists of the following rules:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, 10), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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].

1024
1024_1(x0)
if(true, x0)
if(false, x0)

(25) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, 10), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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

(26) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule 1024_1¹(x) → IF(lt(x, 10), x) we obtained the following new rules [LPAR04]:

1024_1¹(s(z0)) → IF(lt(s(z0), 10), s(z0))

(27) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(s(z0)) → IF(lt(s(z0), 10), s(z0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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

(28) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule 1024_1¹(x) → IF(lt(x, 10), x) at position [0,1] we obtained the following new rules [LPAR04]:

1024_1¹(x) → IF(lt(x, double(s(double(s(s(0)))))), x)

(29) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, double(s(double(s(s(0)))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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:

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, double(s(double(s(s(0)))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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].

10

(33) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, double(s(double(s(s(0)))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(34) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule 1024_1¹(x) → IF(lt(x, double(s(double(s(s(0)))))), x) at position [0,1] we obtained the following new rules [LPAR04]:

1024_1¹(x) → IF(lt(x, s(s(double(double(s(s(0))))))), x)

(35) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, s(s(double(double(s(s(0))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(36) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule 1024_1¹(x) → IF(lt(x, s(s(double(double(s(s(0))))))), x) at position [0,1,0,0,0] we obtained the following new rules [LPAR04]:

1024_1¹(x) → IF(lt(x, s(s(double(s(s(double(s(0)))))))), x)

(37) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, s(s(double(s(s(double(s(0)))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(38) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule 1024_1¹(x) → IF(lt(x, s(s(double(s(s(double(s(0)))))))), x) at position [0,1,0,0] we obtained the following new rules [LPAR04]:

1024_1¹(x) → IF(lt(x, s(s(s(s(double(s(double(s(0))))))))), x)

(39) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, s(s(s(s(double(s(double(s(0))))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(40) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule 1024_1¹(x) → IF(lt(x, s(s(s(s(double(s(double(s(0))))))))), x) at position [0,1,0,0,0,0] we obtained the following new rules [LPAR04]:

1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(double(double(s(0)))))))))), x)

(41) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(double(double(s(0)))))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(42) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule 1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(double(double(s(0)))))))))), x) at position [0,1,0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]:

1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(double(s(s(double(0))))))))))), x)

(43) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(double(s(s(double(0))))))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(44) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule 1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(double(s(s(double(0))))))))))), x) at position [0,1,0,0,0,0,0,0] we obtained the following new rules [LPAR04]:

1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(double(s(double(0)))))))))))), x)

(45) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(double(s(double(0)))))))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(46) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule 1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(double(s(double(0)))))))))))), x) at position [0,1,0,0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]:

1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(double(0))))))))))))), x)

(47) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(double(0))))))))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(48) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule 1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(double(0))))))))))))), x) at position [0,1,0,0,0,0,0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]:

1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(0)))))))))))), x)

(49) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(0)))))))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(50) 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.

(51) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(0)))))))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(52) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule 1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(0)))))))))))), x) at position [0,1,0,0,0,0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]:

1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x)

(53) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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:

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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].

double(0)
double(s(x0))

(57) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x)

The TRS R consists of the following rules:

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

(58) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule 1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x) we obtained the following new rules [LPAR04]:

1024_1¹(s(z0)) → IF(lt(s(z0), s(s(s(s(s(s(s(s(s(s(0))))))))))), s(z0))

(59) Obligation:

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

IF(true, x) → 1024_1¹(s(x))
1024_1¹(s(z0)) → IF(lt(s(z0), s(s(s(s(s(s(s(s(s(s(0))))))))))), s(z0))

The TRS R consists of the following rules:

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

(60) RemovalProof (SOUND transformation)

In the following pairs the term without variables s(s(s(s(s(s(s(s(s(s(0)))))))))) is replaced by the fresh variable x_removed.
Pair: 1024_1¹(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x)
Positions in right side of the pair:

• [0,1]

The new variable was added to all pairs as a new argument[CONREM].

(61) Obligation:

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

IF(true, x, x_removed) → 1024_1¹(s(x), x_removed)
1024_1¹(x, x_removed) → IF(lt(x, x_removed), x, x_removed)

The TRS R consists of the following rules:

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

(62) 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, x, x_removed) → 1024_1¹(s(x), x_removed) the following chains were created:

• We consider the chain 1024_1¹(x, x_removed) → IF(lt(x, x_removed), x, x_removed), IF(true, x, x_removed) → 1024_1¹(s(x), x_removed) which results in the following constraint:
  

  (1)    (IF(lt(x2, x3), x2, x3)=IF(true, x4, x5) ⇒ IF(true, x4, x5)≥1024_1¹(s(x4), x5))

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

  (2)    (lt(x2, x3)=true ⇒ IF(true, x2, x3)≥1024_1¹(s(x2), x3))

  
  
  We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on lt(x2, x3)=true which results in the following new constraints:
  

  (3)    (true=true ⇒ IF(true, 0, s(x12))≥1024_1¹(s(0), s(x12)))

  
  

  (4)    (lt(x14, x13)=true∧(lt(x14, x13)=true ⇒ IF(true, x14, x13)≥1024_1¹(s(x14), x13)) ⇒ IF(true, s(x14), s(x13))≥1024_1¹(s(s(x14)), s(x13)))

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

  (5)    (IF(true, 0, s(x12))≥1024_1¹(s(0), s(x12)))

  
  
  We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (lt(x14, x13)=true ⇒ IF(true, x14, x13)≥1024_1¹(s(x14), x13)) with σ = [ ] which results in the following new constraint:
  

  (6)    (IF(true, x14, x13)≥1024_1¹(s(x14), x13) ⇒ IF(true, s(x14), s(x13))≥1024_1¹(s(s(x14)), s(x13)))

  
  
  

For Pair 1024_1¹(x, x_removed) → IF(lt(x, x_removed), x, x_removed) the following chains were created:

• We consider the chain IF(true, x, x_removed) → 1024_1¹(s(x), x_removed), 1024_1¹(x, x_removed) → IF(lt(x, x_removed), x, x_removed) which results in the following constraint:
  

  (7)    (1024_1¹(s(x6), x7)=1024_1¹(x8, x9) ⇒ 1024_1¹(x8, x9)≥IF(lt(x8, x9), x8, x9))

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

  (8)    (1024_1¹(s(x6), x7)≥IF(lt(s(x6), x7), s(x6), x7))

  
  
  

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

• IF(true, x, x_removed) → 1024_1¹(s(x), x_removed)
  
  • (IF(true, 0, s(x12))≥1024_1¹(s(0), s(x12)))
    
  • (IF(true, x14, x13)≥1024_1¹(s(x14), x13) ⇒ IF(true, s(x14), s(x13))≥1024_1¹(s(s(x14)), s(x13)))
    
  
  
• 1024_1¹(x, x_removed) → IF(lt(x, x_removed), x, x_removed)
  
  • (1024_1¹(s(x6), x7)≥IF(lt(s(x6), x7), s(x6), x7))
    
  
  

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) = 1   
POL(1024_1¹(x₁, x₂)) = 1 - x₁ + x₂   
POL(IF(x₁, x₂, x₃)) = 1 - x₁ - x₂ + x₃   
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, x, x_removed) → 1024_1¹(s(x), x_removed)

The following pairs are in P_bound:

IF(true, x, x_removed) → 1024_1¹(s(x), x_removed)

The following rules are usable:

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

(63) Obligation:

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

1024_1¹(x, x_removed) → IF(lt(x, x_removed), x, x_removed)

The TRS R consists of the following rules:

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

(64) DependencyGraphProof (EQUIVALENT transformation)

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

(65) TRUE