YES Termination w.r.t. Q proof of AProVE_07_thiemann19.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 4 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 UsableRulesProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QReductionProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 TransformationProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 TransformationProof (⇔, 0 ms)
        ↳35 QDP
        ↳36 TransformationProof (⇔, 0 ms)
        ↳37 QDP
        ↳38 TransformationProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 TransformationProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 TransformationProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 TransformationProof (⇔, 0 ms)
        ↳45 QDP
        ↳46 TransformationProof (⇔, 0 ms)
        ↳47 QDP
        ↳48 UsableRulesProof (⇔, 0 ms)
        ↳49 QDP
        ↳50 TransformationProof (⇔, 0 ms)
        ↳51 QDP
        ↳52 UsableRulesProof (⇔, 0 ms)
        ↳53 QDP
        ↳54 QReductionProof (⇔, 0 ms)
        ↳55 QDP
        ↳56 RemovalProof (⇒, 0 ms)
        ↳57 QDP
        ↳58 NonInfProof (⇔, 0 ms)
        ↳59 AND
            ↳60 QDP
                ↳61 DependencyGraphProof (⇔, 0 ms)
                ↳62 TRUE
            ↳63 QDP
                ↳64 DependencyGraphProof (⇔, 0 ms)
                ↳65 TRUE

(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) YES

(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) YES

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

(27) 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.

(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:

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.

(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].

10

(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))

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

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

(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, 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.

(34) TransformationProof (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) → 1024_11(x) → IF(lt(x, s(s(double(s(s(double(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(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.

(36) TransformationProof (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) → 1024_11(x) → IF(lt(x, s(s(s(s(double(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(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.

(38) TransformationProof (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) → 1024_11(x) → IF(lt(x, s(s(s(s(s(s(double(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(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.

(40) TransformationProof (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) → 1024_11(x) → IF(lt(x, s(s(s(s(s(s(double(s(s(double(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(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.

(42) TransformationProof (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) → 1024_11(x) → IF(lt(x, s(s(s(s(s(s(s(s(double(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(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.

(44) TransformationProof (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) → 1024_11(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(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(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.

(46) TransformationProof (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) → 1024_11(x) → IF(lt(x, s(s(s(s(s(s(s(s(s(s(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(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) 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.

(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(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.

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

(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(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) 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.

(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:

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) 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))

(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))

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

(56) 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].

(57) 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.

(58) 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(x2, x3) → IF(lt(x2, x3), x2, x3), IF(true, x4, x5) → 1024_11(s(x4), x5) 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, x6, x7) → 1024_11(s(x6), x7), 1024_11(x8, x9) → IF(lt(x8, x9), x8, x9) which results in the following constraint:
    (1)    (1024_11(s(x6), x7)=1024_11(x8, x9) ⇒ 1024_11(x8, x9)≥IF(lt(x8, x9), x8, x9))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (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 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(1024_11(x1, x2)) = -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>:

1024_11(x, x_removed) → IF(lt(x, x_removed), x, x_removed)
The following pairs are in Pbound:

IF(true, x, x_removed) → 1024_11(s(x), x_removed)
The following rules are usable:

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

(59) Complex Obligation (AND)

(60) Obligation:

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

IF(true, x, x_removed) → 1024_11(s(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.

(61) DependencyGraphProof (EQUIVALENT transformation)

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

(62) TRUE

(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