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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

LE(s(x), s(y)) → LE(x, y)
MINUS(x, s(y)) → PRED(minus(x, y))
MINUS(x, s(y)) → MINUS(x, y)
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
MOD(s(x), s(y)) → LE(y, x)
IF_MOD(true, s(x), s(y)) → MOD(minus(x, y), s(y))
IF_MOD(true, 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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

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

(4) Complex Obligation (AND)

(5) Obligation:

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

MINUS(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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

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

(6) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

• MINUS(x, s(y)) → MINUS(x, y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) 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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

• LE(s(x), s(y)) → LE(x, y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE

(11) Obligation:

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

MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
IF_MOD(true, s(x), s(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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Combined order from the following AFS and order.
minus(x1, x2)  =  x1
0  =  0
s(x1)  =  s(x1)
pred(x1)  =  x1
true  =  true

Homeomorphic Embedding Order

AFS:
minus(x1, x2)  =  x1
0  =  0
s(x1)  =  s(x1)
pred(x1)  =  x1
true  =  true

From the DPs we obtained the following set of size-change graphs:

• IF_MOD(true, s(x), s(y)) → MOD(minus(x, y), s(y)) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 > 1, 3 >= 2

• MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y)) (allowed arguments on rhs = {2, 3})
  The graph contains the following edges 1 >= 2, 2 >= 3

We oriented the following set of usable rules [AAECC05,FROCOS05].

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

(13) TRUE