Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → 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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

(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:

LT(s(x), s(y)) → LT(x, y)
TIMES(s(x), y) → PLUS(y, times(x, y))
TIMES(s(x), y) → TIMES(x, y)
PLUS(s(x), y) → PLUS(x, y)
FAC(x) → LOOP(x, s(0), s(0))
LOOP(x, c, y) → IF(lt(x, c), x, c, y)
LOOP(x, c, y) → LT(x, c)
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
IF(false, x, c, y) → TIMES(y, s(c))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PLUS(s(x), y) → PLUS(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TIMES(s(x), y) → TIMES(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) 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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LT(s(x), s(y)) → LT(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LT(x1, x2)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE

(22) Obligation:

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

LOOP(x, c, y) → IF(lt(x, c), x, c, y)
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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