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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, 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:

LE(s(x), s(y)) → LE(x, y)
ID(s(x)) → ID(x)
MINUS(s(x), s(y)) → MINUS(x, y)
MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y))
MOD(x, y) → ZERO(x)
MOD(x, y) → ZERO(y)
MOD(x, y) → LE(y, x)
MOD(x, y) → ID(x)
MOD(x, y) → ID(y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
IF2(false, b2, x, y) → IF3(b2, x, y)
IF3(true, x, y) → MOD(minus(x, y), s(y))
IF3(true, x, y) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, 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 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, 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:

MINUS(s(x), s(y)) → MINUS(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, 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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

(11) Obligation:

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

MINUS(s(x), s(y)) → MINUS(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:

• MINUS(s(x), s(y)) → MINUS(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:

ID(s(x)) → ID(x)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, 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:

ID(s(x)) → ID(x)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, 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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

(18) Obligation:

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

ID(s(x)) → ID(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:

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

(20) TRUE

(21) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, 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:

LE(s(x), s(y)) → LE(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, 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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

(25) Obligation:

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

LE(s(x), s(y)) → LE(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:

• LE(s(x), s(y)) → LE(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:

MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y))
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
IF2(false, b2, x, y) → IF3(b2, x, y)
IF3(true, x, y) → MOD(minus(x, y), s(y))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, 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:

MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y))
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
IF2(false, b2, x, y) → IF3(b2, x, y)
IF3(true, x, y) → MOD(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, 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, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

(32) Obligation:

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

MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y))
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
IF2(false, b2, x, y) → IF3(b2, x, y)
IF3(true, x, y) → MOD(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(33) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y)) we obtained the following new rules [LPAR04]:

MOD(y_0, s(z1)) → IF_MOD(zero(y_0), zero(s(z1)), le(s(z1), y_0), id(y_0), id(s(z1)))

(34) Obligation:

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

IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
IF2(false, b2, x, y) → IF3(b2, x, y)
IF3(true, x, y) → MOD(minus(x, y), s(y))
MOD(y_0, s(z1)) → IF_MOD(zero(y_0), zero(s(z1)), le(s(z1), y_0), id(y_0), id(s(z1)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(35) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y)) at position [0] we obtained the following new rules [LPAR04]:

MOD(0, y1) → IF_MOD(true, zero(y1), le(y1, 0), id(0), id(y1))
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), id(s(x0)), id(y1))

(36) Obligation:

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

IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
IF2(false, b2, x, y) → IF3(b2, x, y)
IF3(true, x, y) → MOD(minus(x, y), s(y))
MOD(0, y1) → IF_MOD(true, zero(y1), le(y1, 0), id(0), id(y1))
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), id(s(x0)), id(y1))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(37) DependencyGraphProof (EQUIVALENT transformation)

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

(38) Obligation:

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

IF2(false, b2, x, y) → IF3(b2, x, y)
IF3(true, x, y) → MOD(minus(x, y), s(y))
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), id(s(x0)), id(y1))
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(39) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), id(s(x0)), id(y1)) at position [3] we obtained the following new rules [LPAR04]:

MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1))

(40) Obligation:

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

IF2(false, b2, x, y) → IF3(b2, x, y)
IF3(true, x, y) → MOD(minus(x, y), s(y))
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(41) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y) we obtained the following new rules [LPAR04]:

IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)

(42) Obligation:

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

IF2(false, b2, x, y) → IF3(b2, x, y)
IF3(true, x, y) → MOD(minus(x, y), s(y))
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(43) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF3(true, x, y) → MOD(minus(x, y), s(y)) at position [0] we obtained the following new rules [LPAR04]:

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))

(44) Obligation:

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

IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1))
IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(45) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y) we obtained the following new rules [LPAR04]:

IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)

(46) Obligation:

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

IF2(false, b2, x, y) → IF3(b2, x, y)
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1))
IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(47) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF2(false, b2, x, y) → IF3(b2, x, y) we obtained the following new rules [LPAR04]:

IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)

(48) Obligation:

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

MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1))
IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(49) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1)) we obtained the following new rules [LPAR04]:

MOD(s(x0), s(0)) → IF_MOD(false, zero(s(0)), le(s(0), s(x0)), s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, zero(s(s(z1))), le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))

(50) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(0)) → IF_MOD(false, zero(s(0)), le(s(0), s(x0)), s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, zero(s(s(z1))), le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(51) 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.

(52) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(0)) → IF_MOD(false, zero(s(0)), le(s(0), s(x0)), s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, zero(s(s(z1))), le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))

The TRS R consists of the following rules:

zero(s(x)) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(53) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule MOD(s(x0), s(0)) → IF_MOD(false, zero(s(0)), le(s(0), s(x0)), s(id(x0)), id(s(0))) at position [1] we obtained the following new rules [LPAR04]:

MOD(s(x0), s(0)) → IF_MOD(false, false, le(s(0), s(x0)), s(id(x0)), id(s(0)))

(54) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(s(z1))) → IF_MOD(false, zero(s(s(z1))), le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, le(s(0), s(x0)), s(id(x0)), id(s(0)))

The TRS R consists of the following rules:

zero(s(x)) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(55) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule MOD(s(x0), s(s(z1))) → IF_MOD(false, zero(s(s(z1))), le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1)))) at position [1] we obtained the following new rules [LPAR04]:

MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))

(56) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(0)) → IF_MOD(false, false, le(s(0), s(x0)), s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))

The TRS R consists of the following rules:

zero(s(x)) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(57) 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.

(58) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(0)) → IF_MOD(false, false, le(s(0), s(x0)), s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(59) 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].

zero(0)
zero(s(x0))

(60) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(0)) → IF_MOD(false, false, le(s(0), s(x0)), s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(61) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule MOD(s(x0), s(0)) → IF_MOD(false, false, le(s(0), s(x0)), s(id(x0)), id(s(0))) at position [2] we obtained the following new rules [LPAR04]:

MOD(s(x0), s(0)) → IF_MOD(false, false, le(0, x0), s(id(x0)), id(s(0)))

(62) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, le(0, x0), s(id(x0)), id(s(0)))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(63) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1)))) at position [2] we obtained the following new rules [LPAR04]:

MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), id(s(s(z1))))

(64) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(0)) → IF_MOD(false, false, le(0, x0), s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), id(s(s(z1))))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(65) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule MOD(s(x0), s(0)) → IF_MOD(false, false, le(0, x0), s(id(x0)), id(s(0))) at position [2] we obtained the following new rules [LPAR04]:

MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), id(s(0)))

(66) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), id(s(s(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), id(s(0)))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(67) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), id(s(s(z1)))) at position [4] we obtained the following new rules [LPAR04]:

MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(id(s(z1))))

(68) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(id(s(z1))))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(69) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), id(s(0))) at position [4] we obtained the following new rules [LPAR04]:

MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(id(0)))

(70) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(id(s(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(id(0)))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(71) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(id(s(z1)))) at position [4,0] we obtained the following new rules [LPAR04]:

MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))

(72) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(id(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(73) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(id(0))) at position [4,0] we obtained the following new rules [LPAR04]:

MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(0))

(74) Obligation:

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

IF3(true, x0, 0) → MOD(x0, s(0))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(0))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(75) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF3(true, x0, 0) → MOD(x0, s(0)) we obtained the following new rules [LPAR04]:

IF3(true, s(z1), 0) → MOD(s(z1), s(0))

(76) Obligation:

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

IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(0))
IF3(true, s(z1), 0) → MOD(s(z1), s(0))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(77) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3) we obtained the following new rules [LPAR04]:

IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF_MOD(false, false, true, s(y_0), s(0)) → IF2(false, true, s(y_0), s(0))

(78) Obligation:

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

IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(0))
IF3(true, s(z1), 0) → MOD(s(z1), s(0))
IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF_MOD(false, false, true, s(y_0), s(0)) → IF2(false, true, s(y_0), s(0))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(79) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3) we obtained the following new rules [LPAR04]:

IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2)))
IF2(false, true, s(z0), s(0)) → IF3(true, s(z0), s(0))

(80) Obligation:

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

IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(0))
IF3(true, s(z1), 0) → MOD(s(z1), s(0))
IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF_MOD(false, false, true, s(y_0), s(0)) → IF2(false, true, s(y_0), s(0))
IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2)))
IF2(false, true, s(z0), s(0)) → IF3(true, s(z0), s(0))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(81) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(82) Obligation:

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

MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2)))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(83) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1))) we obtained the following new rules [LPAR04]:

IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))

(84) Obligation:

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

MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2)))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(85) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1)))) we obtained the following new rules [LPAR04]:

MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(id(s(z1)))))

(86) Obligation:

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

IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2)))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(id(s(z1)))))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(87) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(id(s(z1))))) at position [4,0,0] we obtained the following new rules [LPAR04]:

MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(s(id(z1)))))

(88) Obligation:

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

IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2)))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(s(id(z1)))))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(89) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2))) we obtained the following new rules [LPAR04]:

IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))

(90) Obligation:

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

IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(s(id(z1)))))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(91) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2))) we obtained the following new rules [LPAR04]:

IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))

(92) Obligation:

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

IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(s(id(z1)))))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(93) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(s(id(z1))))) at position [2] we obtained the following new rules [LPAR04]:

MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(id(s(x1))), s(s(s(id(y1)))))
MOD(s(0), s(s(s(y1)))) → IF_MOD(false, false, false, s(id(0)), s(s(s(id(y1)))))

(94) Obligation:

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

IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))
MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(id(s(x1))), s(s(s(id(y1)))))
MOD(s(0), s(s(s(y1)))) → IF_MOD(false, false, false, s(id(0)), s(s(s(id(y1)))))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(95) DependencyGraphProof (EQUIVALENT transformation)

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

(96) Obligation:

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

MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(id(s(x1))), s(s(s(id(y1)))))
IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(97) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(id(s(x1))), s(s(s(id(y1))))) at position [3,0] we obtained the following new rules [LPAR04]:

MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(s(id(x1))), s(s(s(id(y1)))))

(98) Obligation:

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

IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(s(id(x1))), s(s(s(id(y1)))))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(99) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(IF2(x₁, x₂, x₃, x₄)) = x₃   
POL(IF3(x₁, x₂, x₃)) = x₂   
POL(IF_MOD(x₁, x₂, x₃, x₄, x₅)) = x₄   
POL(MOD(x₁, x₂)) = x₁   
POL(false) = 0   
POL(id(x₁)) = x₁   
POL(le(x₁, x₂)) = 0   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

(100) Obligation:

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

IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(s(id(x1))), s(s(s(id(y1)))))

The TRS R consists of the following rules:

le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))

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

(101) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

(102) TRUE