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:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(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:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(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:

MINUS(s(x), y) → IF(gt(s(x), y), x, y)
MINUS(s(x), y) → GT(s(x), y)
IF(true, x, y) → MINUS(x, y)
MOD(x, s(y)) → IF1(lt(x, s(y)), x, s(y))
MOD(x, s(y)) → LT(x, s(y))
IF1(false, x, y) → MOD(minus(x, y), y)
IF1(false, x, y) → MINUS(x, y)
GT(s(x), s(y)) → GT(x, y)
LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(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 4 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) 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:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(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:

LT(s(x), s(y)) → LT(x, y)

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

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(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].

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

(11) 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.

(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:

• LT(s(x), s(y)) → LT(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:

GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(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:

GT(s(x), s(y)) → GT(x, y)

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

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(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].

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

(18) Obligation:

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

GT(s(x), s(y)) → GT(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:

• GT(s(x), s(y)) → GT(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, y) → MINUS(x, y)
MINUS(s(x), y) → IF(gt(s(x), y), x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(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:

IF(true, x, y) → MINUS(x, y)
MINUS(s(x), y) → IF(gt(s(x), y), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(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].

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

(25) Obligation:

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

IF(true, x, y) → MINUS(x, y)
MINUS(s(x), y) → IF(gt(s(x), y), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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:

• MINUS(s(x), y) → IF(gt(s(x), y), x, y)
  The graph contains the following edges 1 > 2, 2 >= 3

• IF(true, x, y) → MINUS(x, y)
  The graph contains the following edges 2 >= 1, 3 >= 2

(27) YES

(28) Obligation:

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

IF1(false, x, y) → MOD(minus(x, y), y)
MOD(x, s(y)) → IF1(lt(x, s(y)), x, s(y))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(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:

IF1(false, x, y) → MOD(minus(x, y), y)
MOD(x, s(y)) → IF1(lt(x, s(y)), x, s(y))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)
lt(x, 0) → false
minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(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].

mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)

(32) Obligation:

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

IF1(false, x, y) → MOD(minus(x, y), y)
MOD(x, s(y)) → IF1(lt(x, s(y)), x, s(y))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)
lt(x, 0) → false
minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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

(33) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule MOD(x, s(y)) → IF1(lt(x, s(y)), x, s(y)) at position [0] we obtained the following new rules [LPAR04]:

MOD(0, s(x0)) → IF1(true, 0, s(x0))
MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))

(34) Obligation:

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

IF1(false, x, y) → MOD(minus(x, y), y)
MOD(0, s(x0)) → IF1(true, 0, s(x0))
MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)
lt(x, 0) → false
minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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

(35) DependencyGraphProof (EQUIVALENT transformation)

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

(36) Obligation:

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

MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))
IF1(false, x, y) → MOD(minus(x, y), y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)
lt(x, 0) → false
minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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

(37) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF1(false, x, y) → MOD(minus(x, y), y) at position [0] we obtained the following new rules [LPAR04]:

IF1(false, 0, x0) → MOD(0, x0)
IF1(false, s(x0), x1) → MOD(if(gt(s(x0), x1), x0, x1), x1)

(38) Obligation:

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

MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))
IF1(false, 0, x0) → MOD(0, x0)
IF1(false, s(x0), x1) → MOD(if(gt(s(x0), x1), x0, x1), x1)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)
lt(x, 0) → false
minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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

(39) DependencyGraphProof (EQUIVALENT transformation)

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

(40) Obligation:

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

IF1(false, s(x0), x1) → MOD(if(gt(s(x0), x1), x0, x1), x1)
MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)
lt(x, 0) → false
minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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

(41) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF1(false, s(x0), x1) → MOD(if(gt(s(x0), x1), x0, x1), x1) we obtained the following new rules [LPAR04]:

IF1(false, s(z0), s(z1)) → MOD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1))

(42) Obligation:

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

MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))
IF1(false, s(z0), s(z1)) → MOD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)
lt(x, 0) → false
minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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

(43) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF1(false, s(z0), s(z1)) → MOD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1)) at position [0,0] we obtained the following new rules [LPAR04]:

IF1(false, s(z0), s(z1)) → MOD(if(gt(z0, z1), z0, s(z1)), s(z1))

(44) Obligation:

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

MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))
IF1(false, s(z0), s(z1)) → MOD(if(gt(z0, z1), z0, s(z1)), s(z1))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)
lt(x, 0) → false
minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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

(45) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
IF1(false, s(z0), s(z1)) → MOD(if(gt(z0, z1), z0, s(z1)), s(z1))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(IF1(x₁, x₂, x₃)) = x₂   
POL(MOD(x₁, x₂)) = x₁   
POL(false) = 1   
POL(gt(x₁, x₂)) = 1   
POL(if(x₁, x₂, x₃)) = x₁ + x₂   
POL(lt(x₁, x₂)) = 1   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = 1 + x₁   
POL(true) = 1   

At least one of these decreasing rules is always used after the deleted DP:
if(false, x683, y573) → 0


The following formula is valid:
z0:sort[a22],z1:sort[a22].if'(gt(z0 , z1 ), z0 , s(z1 ))=true


The transformed set:
minus'(0, y18) → false
minus'(s(x32), y26) → if'(gt(s(x32), y26), x32, y26)
if'(true, x59, y49) → minus'(x59, y49)
if'(false, x68, y57) → true
lt(0, s(x)) → true
lt(s(x6), s(y3)) → lt(x6, y3)
lt(x15, 0) → false
minus(0, y18) → 0
minus(s(x32), y26) → if(gt(s(x32), y26), x32, y26)
gt(s(x41), 0) → true
gt(s(x50), s(y41)) → gt(x50, y41)
if(true, x59, y49) → s(minus(x59, y49))
if(false, x68, y57) → 0
gt(0, y65) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a22](0, 0) → true
equal_sort[a22](0, s(x0)) → false
equal_sort[a22](s(x0), 0) → false
equal_sort[a22](s(x0), s(x1)) → equal_sort[a22](x0, x1)
equal_sort[a39](witness_sort[a39], witness_sort[a39]) → true


The proof given by the theorem prover:
The following program was given to the internal theorem prover:

   [x, x0, x1, y18, x32, y26, x68, y57, y49, x6, y3, x15, x41, x50, y41, y65]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a22](0, 0) -> true
   equal_sort[a22](0, s(x0)) -> false
   equal_sort[a22](s(x0), 0) -> false
   equal_sort[a22](s(x0), s(x1)) -> equal_sort[a22](x0, x1)
   equal_sort[a39](witness_sort[a39], witness_sort[a39]) -> true
   minus'(0, y18) -> false
   equal_bool(gt(s(x32), y26), true) -> true | minus'(s(x32), y26) -> minus'(x32, y26)
   equal_bool(gt(s(x32), y26), true) -> false | minus'(s(x32), y26) -> true
   if'(false, x68, y57) -> true
   if'(true, 0, y49) -> false
   if'(true, s(x32), y49) -> if'(gt(s(x32), y49), x32, y49)
   lt(0, s(x)) -> true
   lt(s(x6), s(y3)) -> lt(x6, y3)
   lt(x15, 0) -> false
   minus(0, y18) -> 0
   equal_bool(gt(s(x32), y26), true) -> true | minus(s(x32), y26) -> s(minus(x32, y26))
   equal_bool(gt(s(x32), y26), true) -> false | minus(s(x32), y26) -> 0
   gt(s(x41), 0) -> true
   gt(s(x50), s(y41)) -> gt(x50, y41)
   gt(0, y65) -> false
   if(false, x68, y57) -> 0
   if(true, 0, y49) -> s(0)
   if(true, s(x32), y49) -> s(if(gt(s(x32), y49), x32, y49))

The following output was given by the internal theorem prover:

proof of internal
# AProVE Commit ID: 9a00b172b26c9abb2d4c4d5eaf341e919eb0fbf1 nowonder 20100222 unpublished dirty


Partial correctness of the following Program

   [x, x0, x1, y18, x32, y26, x68, y57, y49, x6, y3, x15, x41, x50, y41, y65]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a22](0, 0) -> true
   equal_sort[a22](0, s(x0)) -> false
   equal_sort[a22](s(x0), 0) -> false
   equal_sort[a22](s(x0), s(x1)) -> equal_sort[a22](x0, x1)
   equal_sort[a39](witness_sort[a39], witness_sort[a39]) -> true
   minus'(0, y18) -> false
   equal_bool(gt(s(x32), y26), true) -> true | minus'(s(x32), y26) -> minus'(x32, y26)
   equal_bool(gt(s(x32), y26), true) -> false | minus'(s(x32), y26) -> true
   if'(false, x68, y57) -> true
   if'(true, 0, y49) -> false
   if'(true, s(x32), y49) -> if'(gt(s(x32), y49), x32, y49)
   lt(0, s(x)) -> true
   lt(s(x6), s(y3)) -> lt(x6, y3)
   lt(x15, 0) -> false
   minus(0, y18) -> 0
   equal_bool(gt(s(x32), y26), true) -> true | minus(s(x32), y26) -> s(minus(x32, y26))
   equal_bool(gt(s(x32), y26), true) -> false | minus(s(x32), y26) -> 0
   gt(s(x41), 0) -> true
   gt(s(x50), s(y41)) -> gt(x50, y41)
   gt(0, y65) -> false
   if(false, x68, y57) -> 0
   if(true, 0, y49) -> s(0)
   if(true, s(x32), y49) -> s(if(gt(s(x32), y49), x32, y49))

using the following formula:
z0:sort[a22],z1:sort[a22].if'(gt(z0, z1), z0, s(z1))=true

could be successfully shown:
(0) Formula
(1) Induction by algorithm [EQUIVALENT]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT]
        (5) Formula
        (6) Induction by data structure [EQUIVALENT]
        (7) AND
            (8) Formula
                (9) Symbolic evaluation [EQUIVALENT]
                (10) YES
            (11) Formula
                (12) Symbolic evaluation under hypothesis [EQUIVALENT]
                (13) YES
    (14) Formula
        (15) Symbolic evaluation [EQUIVALENT]
        (16) YES
    (17) Formula
        (18) Symbolic evaluation [EQUIVALENT]
        (19) Formula
        (20) Hypothesis Lifting [EQUIVALENT]
        (21) Formula
        (22) Inverse Substitution [SOUND]
        (23) Formula
        (24) Inverse Substitution [SOUND]
        (25) Formula
        (26) Induction by algorithm [EQUIVALENT]
        (27) AND
            (28) Formula
                (29) Symbolic evaluation [EQUIVALENT]
                (30) YES
            (31) Formula
                (32) Symbolic evaluation [EQUIVALENT]
                (33) YES
            (34) Formula
                (35) Symbolic evaluation under hypothesis [EQUIVALENT]
                (36) YES


----------------------------------------

(0)
Obligation:
Formula:
z0:sort[a22],z1:sort[a22].if'(gt(z0, z1), z0, s(z1))=true

There are no hypotheses.




----------------------------------------

(1) Induction by algorithm (EQUIVALENT)
Induction by algorithm gt(z0, z1) generates the following cases:

1. Base Case:
Formula:
x41:sort[a22].if'(gt(s(x41), 0), s(x41), s(0))=true

There are no hypotheses.





2. Base Case:
Formula:
y65:sort[a22].if'(gt(0, y65), 0, s(y65))=true

There are no hypotheses.





1. Step Case:
Formula:
x50:sort[a22],y41:sort[a22].if'(gt(s(x50), s(y41)), s(x50), s(s(y41)))=true

Hypotheses:
x50:sort[a22],y41:sort[a22].if'(gt(x50, y41), x50, s(y41))=true






----------------------------------------

(2)
Complex Obligation (AND)

----------------------------------------

(3)
Obligation:
Formula:
x41:sort[a22].if'(gt(s(x41), 0), s(x41), s(0))=true

There are no hypotheses.




----------------------------------------

(4) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x41:sort[a22].if'(gt(x41, 0), x41, s(0))=true
----------------------------------------

(5)
Obligation:
Formula:
x41:sort[a22].if'(gt(x41, 0), x41, s(0))=true

There are no hypotheses.




----------------------------------------

(6) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a22] generates the following cases:



1. Base Case:
Formula:
if'(gt(0, 0), 0, s(0))=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a22].if'(gt(s(n), 0), s(n), s(0))=true

Hypotheses:
n:sort[a22].if'(gt(n, 0), n, s(0))=true






----------------------------------------

(7)
Complex Obligation (AND)

----------------------------------------

(8)
Obligation:
Formula:
if'(gt(0, 0), 0, s(0))=true

There are no hypotheses.




----------------------------------------

(9) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(10)
YES

----------------------------------------

(11)
Obligation:
Formula:
n:sort[a22].if'(gt(s(n), 0), s(n), s(0))=true

Hypotheses:
n:sort[a22].if'(gt(n, 0), n, s(0))=true




----------------------------------------

(12) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n:sort[a22].if'(gt(n, 0), n, s(0))=true

----------------------------------------

(13)
YES

----------------------------------------

(14)
Obligation:
Formula:
y65:sort[a22].if'(gt(0, y65), 0, s(y65))=true

There are no hypotheses.




----------------------------------------

(15) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(16)
YES

----------------------------------------

(17)
Obligation:
Formula:
x50:sort[a22],y41:sort[a22].if'(gt(s(x50), s(y41)), s(x50), s(s(y41)))=true

Hypotheses:
x50:sort[a22],y41:sort[a22].if'(gt(x50, y41), x50, s(y41))=true




----------------------------------------

(18) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x50:sort[a22],y41:sort[a22].if'(gt(x50, y41), s(x50), s(s(y41)))=true
----------------------------------------

(19)
Obligation:
Formula:
x50:sort[a22],y41:sort[a22].if'(gt(x50, y41), s(x50), s(s(y41)))=true

Hypotheses:
x50:sort[a22],y41:sort[a22].if'(gt(x50, y41), x50, s(y41))=true




----------------------------------------

(20) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
x50:sort[a22],y41:sort[a22].(if'(gt(x50, y41), x50, s(y41))=true->if'(gt(x50, y41), s(x50), s(s(y41)))=true)

There are no hypotheses.




----------------------------------------

(21)
Obligation:
Formula:
x50:sort[a22],y41:sort[a22].(if'(gt(x50, y41), x50, s(y41))=true->if'(gt(x50, y41), s(x50), s(s(y41)))=true)

There are no hypotheses.




----------------------------------------

(22) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:bool,x50:sort[a22],y41:sort[a22].(if'(n, x50, s(y41))=true->if'(n, s(x50), s(s(y41)))=true)

Inverse substitution used:
[gt(x50, y41)/n]


----------------------------------------

(23)
Obligation:
Formula:
n:bool,x50:sort[a22],y41:sort[a22].(if'(n, x50, s(y41))=true->if'(n, s(x50), s(s(y41)))=true)

There are no hypotheses.




----------------------------------------

(24) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:bool,x50:sort[a22],n':sort[a22].(if'(n, x50, n')=true->if'(n, s(x50), s(n'))=true)

Inverse substitution used:
[s(y41)/n']


----------------------------------------

(25)
Obligation:
Formula:
n:bool,x50:sort[a22],n':sort[a22].(if'(n, x50, n')=true->if'(n, s(x50), s(n'))=true)

There are no hypotheses.




----------------------------------------

(26) Induction by algorithm (EQUIVALENT)
Induction by algorithm if'(n, x50, n') generates the following cases:

1. Base Case:
Formula:
x68:sort[a22],y57:sort[a22].(if'(false, x68, y57)=true->if'(false, s(x68), s(y57))=true)

There are no hypotheses.





2. Base Case:
Formula:
y49:sort[a22].(if'(true, 0, y49)=true->if'(true, s(0), s(y49))=true)

There are no hypotheses.





1. Step Case:
Formula:
x32:sort[a22],y49:sort[a22].(if'(true, s(x32), y49)=true->if'(true, s(s(x32)), s(y49))=true)

Hypotheses:
x32:sort[a22],y49:sort[a22].(if'(gt(s(x32), y49), x32, y49)=true->if'(gt(s(x32), y49), s(x32), s(y49))=true)






----------------------------------------

(27)
Complex Obligation (AND)

----------------------------------------

(28)
Obligation:
Formula:
x68:sort[a22],y57:sort[a22].(if'(false, x68, y57)=true->if'(false, s(x68), s(y57))=true)

There are no hypotheses.




----------------------------------------

(29) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(30)
YES

----------------------------------------

(31)
Obligation:
Formula:
y49:sort[a22].(if'(true, 0, y49)=true->if'(true, s(0), s(y49))=true)

There are no hypotheses.




----------------------------------------

(32) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(33)
YES

----------------------------------------

(34)
Obligation:
Formula:
x32:sort[a22],y49:sort[a22].(if'(true, s(x32), y49)=true->if'(true, s(s(x32)), s(y49))=true)

Hypotheses:
x32:sort[a22],y49:sort[a22].(if'(gt(s(x32), y49), x32, y49)=true->if'(gt(s(x32), y49), s(x32), s(y49))=true)




----------------------------------------

(35) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

x32:sort[a22],y49:sort[a22].(if'(gt(s(x32), y49), x32, y49)=true->if'(gt(s(x32), y49), s(x32), s(y49))=true)

----------------------------------------

(36)
YES

(46) Complex Obligation (AND)

(47) Obligation:

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

MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)
lt(x, 0) → false
minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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

(48) DependencyGraphProof (EQUIVALENT transformation)

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

(49) TRUE

(50) Obligation:

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

minus'(0, y18) → false
minus'(s(x32), y26) → if'(gt(s(x32), y26), x32, y26)
if'(true, x59, y49) → minus'(x59, y49)
if'(false, x68, y57) → true
lt(0, s(x)) → true
lt(s(x6), s(y3)) → lt(x6, y3)
lt(x15, 0) → false
minus(0, y18) → 0
minus(s(x32), y26) → if(gt(s(x32), y26), x32, y26)
gt(s(x41), 0) → true
gt(s(x50), s(y41)) → gt(x50, y41)
if(true, x59, y49) → s(minus(x59, y49))
if(false, x68, y57) → 0
gt(0, y65) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a22](0, 0) → true
equal_sort[a22](0, s(x0)) → false
equal_sort[a22](s(x0), 0) → false
equal_sort[a22](s(x0), s(x1)) → equal_sort[a22](x0, x1)
equal_sort[a39](witness_sort[a39], witness_sort[a39]) → true

Q is empty.

(51) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
minus'(x1, x2)  =  minus'(x1, x2)
0  =  0
false  =  false
s(x1)  =  s(x1)
if'(x1, x2, x3)  =  if'(x1, x2, x3)
gt(x1, x2)  =  gt(x1, x2)
true  =  true
lt(x1, x2)  =  lt(x1, x2)
minus(x1, x2)  =  minus(x1, x2)
if(x1, x2, x3)  =  if(x1, x2, x3)
equal_bool(x1, x2)  =  equal_bool(x1, x2)
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)
not(x1)  =  not(x1)
isa_true(x1)  =  isa_true(x1)
isa_false(x1)  =  x1
equal_sort[a22](x1, x2)  =  equal_sort[a22](x1, x2)
equal_sort[a39](x1, x2)  =  equal_sort[a39](x1, x2)
witness_sort[a39]  =  witness_sort[a39]

Recursive path order with status [RPO].
Quasi-Precedence:

[minus'₂, 0, false, if'₃, true, minus₂, if₃, isa_true1, equal_sort[a39]₂] > s₁ > lt₂
[minus'₂, 0, false, if'₃, true, minus₂, if₃, isa_true1, equal_sort[a39]₂] > s₁ > equal_sort[a22]₂
[minus'₂, 0, false, if'₃, true, minus₂, if₃, isa_true1, equal_sort[a39]₂] > gt₂

Status:

minus'₂: [2,1]
0: multiset
false: multiset
s₁: [1]
if'₃: [3,2,1]
gt₂: [2,1]
true: multiset
lt₂: [2,1]
minus₂: [2,1]
if₃: [3,2,1]
equal_bool2: multiset
and₂: [2,1]
or₂: multiset
not₁: multiset
isa_true1: [1]
equal_sort[a22]₂: multiset
equal_sort[a39]₂: multiset
witness_sort[a39]: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

minus'(0, y18) → false
minus'(s(x32), y26) → if'(gt(s(x32), y26), x32, y26)
if'(true, x59, y49) → minus'(x59, y49)
if'(false, x68, y57) → true
lt(0, s(x)) → true
lt(s(x6), s(y3)) → lt(x6, y3)
lt(x15, 0) → false
minus(0, y18) → 0
minus(s(x32), y26) → if(gt(s(x32), y26), x32, y26)
gt(s(x41), 0) → true
gt(s(x50), s(y41)) → gt(x50, y41)
if(true, x59, y49) → s(minus(x59, y49))
if(false, x68, y57) → 0
gt(0, y65) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
equal_sort[a22](0, 0) → true
equal_sort[a22](0, s(x0)) → false
equal_sort[a22](s(x0), 0) → false
equal_sort[a22](s(x0), s(x1)) → equal_sort[a22](x0, x1)
equal_sort[a39](witness_sort[a39], witness_sort[a39]) → true

(52) Obligation:

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

isa_false(true) → false
isa_false(false) → true

Q is empty.

(53) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(false) = 2   
POL(isa_false(x₁)) = 2 + 2·x₁   
POL(true) = 1   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

isa_false(true) → false
isa_false(false) → true

(54) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(55) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(56) YES