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(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))

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(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

(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)
PLUS(s(x), y) → PLUS(x, y)
TIMES(s(x), y) → PLUS(y, times(x, y))
TIMES(s(x), y) → TIMES(x, y)
LOG(s(x), s(s(b))) → LOOP(s(x), s(s(b)), s(0), 0)
LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), s(y), z)
LOOP(x, s(s(b)), s(y), z) → LE(x, s(y))
IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))
IF(false, x, b, y, z) → TIMES(b, y)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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:

PLUS(s(x), y) → PLUS(x, y)

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

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

(11) Obligation:

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

PLUS(s(x), y) → PLUS(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:

• PLUS(s(x), y) → PLUS(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:

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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:

TIMES(s(x), y) → TIMES(x, y)

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

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

(18) Obligation:

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

TIMES(s(x), y) → TIMES(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:

• TIMES(s(x), y) → TIMES(x, y)
  The graph contains the following edges 1 > 1, 2 >= 2

(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(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

(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:

LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), s(y), z)
IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))

The TRS R consists of the following rules:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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:

LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), s(y), z)
IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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].

log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

(32) Obligation:

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

LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), s(y), z)
IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(33) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), s(y), z) we obtained the following new rules [LPAR04]:

LOOP(z0, s(s(x1)), s(x2), s(z3)) → IF(le(z0, s(x2)), z0, s(s(x1)), s(x2), s(z3))

(34) Obligation:

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

IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))
LOOP(z0, s(s(x1)), s(x2), s(z3)) → IF(le(z0, s(x2)), z0, s(s(x1)), s(x2), s(z3))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(35) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), s(y), z) at position [0] we obtained the following new rules [LPAR04]:

LOOP(0, s(s(y1)), s(y2), y3) → IF(true, 0, s(s(y1)), s(y2), y3)
LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)

(36) Obligation:

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

IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))
LOOP(0, s(s(y1)), s(y2), y3) → IF(true, 0, s(s(y1)), s(y2), y3)
LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), 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:

LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(39) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z)) at position [2] we obtained the following new rules [LPAR04]:

IF(false, y0, 0, x0, y3) → LOOP(y0, 0, 0, s(y3))
IF(false, y0, s(x0), x1, y3) → LOOP(y0, s(x0), plus(x1, times(x0, x1)), s(y3))

(40) Obligation:

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

LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
IF(false, y0, 0, x0, y3) → LOOP(y0, 0, 0, s(y3))
IF(false, y0, s(x0), x1, y3) → LOOP(y0, s(x0), plus(x1, times(x0, x1)), s(y3))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(41) DependencyGraphProof (EQUIVALENT transformation)

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

(42) Obligation:

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

IF(false, y0, s(x0), x1, y3) → LOOP(y0, s(x0), plus(x1, times(x0, x1)), s(y3))
LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(43) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, y0, s(x0), x1, y3) → LOOP(y0, s(x0), plus(x1, times(x0, x1)), s(y3)) we obtained the following new rules [LPAR04]:

IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), plus(s(z2), times(s(z1), s(z2))), s(z3))

(44) Obligation:

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

LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), plus(s(z2), times(s(z1), s(z2))), s(z3))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(45) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), plus(s(z2), times(s(z1), s(z2))), s(z3)) at position [2] we obtained the following new rules [LPAR04]:

IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, times(s(z1), s(z2)))), s(z3))

(46) Obligation:

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

LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, times(s(z1), s(z2)))), s(z3))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(47) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, times(s(z1), s(z2)))), s(z3)) at position [2,0,1] we obtained the following new rules [LPAR04]:

IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, plus(s(z2), times(z1, s(z2))))), s(z3))

(48) Obligation:

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

LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, plus(s(z2), times(z1, s(z2))))), s(z3))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(49) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, plus(s(z2), times(z1, s(z2))))), s(z3)) at position [2,0,1] we obtained the following new rules [LPAR04]:

IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(z3))

(50) Obligation:

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

LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(z3))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(51) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3) we obtained the following new rules [LPAR04]:

LOOP(s(z0), s(s(z1)), s(y_2), s(z3)) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(z3))

(52) Obligation:

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

IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(z3))
LOOP(s(z0), s(s(z1)), s(y_2), s(z3)) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(z3))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(53) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(z3)) we obtained the following new rules [LPAR04]:

IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3)))

(54) Obligation:

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

LOOP(s(z0), s(s(z1)), s(y_2), s(z3)) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(z3))
IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(55) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule LOOP(s(z0), s(s(z1)), s(y_2), s(z3)) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(z3)) we obtained the following new rules [LPAR04]:

LOOP(s(z0), s(s(z1)), s(y_2), s(s(z3))) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(s(z3)))

(56) Obligation:

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

IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3)))
LOOP(s(z0), s(s(z1)), s(y_2), s(s(z3))) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(s(z3)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(57) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3))) the following chains were created:

• We consider the chain LOOP(s(z0), s(s(z1)), s(y_2), s(s(z3))) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(s(z3))), IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3))) which results in the following constraint:
  

  (1)    (IF(le(x4, x6), s(x4), s(s(x5)), s(x6), s(s(x7)))=IF(false, s(x8), s(s(x9)), s(x10), s(x11)) ⇒ IF(false, s(x8), s(s(x9)), s(x10), s(x11))≥LOOP(s(x8), s(s(x9)), s(plus(x10, s(plus(x10, times(x9, s(x10)))))), s(s(x11))))

  
  
  We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
  

  (2)    (le(x4, x6)=false ⇒ IF(false, s(x4), s(s(x5)), s(x6), s(s(x7)))≥LOOP(s(x4), s(s(x5)), s(plus(x6, s(plus(x6, times(x5, s(x6)))))), s(s(s(x7)))))

  
  
  We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x4, x6)=false which results in the following new constraints:
  

  (3)    (le(x26, x25)=false∧(∀x27,x28:le(x26, x25)=false ⇒ IF(false, s(x26), s(s(x27)), s(x25), s(s(x28)))≥LOOP(s(x26), s(s(x27)), s(plus(x25, s(plus(x25, times(x27, s(x25)))))), s(s(s(x28))))) ⇒ IF(false, s(s(x26)), s(s(x5)), s(s(x25)), s(s(x7)))≥LOOP(s(s(x26)), s(s(x5)), s(plus(s(x25), s(plus(s(x25), times(x5, s(s(x25))))))), s(s(s(x7)))))

  
  

  (4)    (false=false ⇒ IF(false, s(s(x29)), s(s(x5)), s(0), s(s(x7)))≥LOOP(s(s(x29)), s(s(x5)), s(plus(0, s(plus(0, times(x5, s(0)))))), s(s(s(x7)))))

  
  
  We simplified constraint (3) using rule (VI) where we applied the induction hypothesis (∀x27,x28:le(x26, x25)=false ⇒ IF(false, s(x26), s(s(x27)), s(x25), s(s(x28)))≥LOOP(s(x26), s(s(x27)), s(plus(x25, s(plus(x25, times(x27, s(x25)))))), s(s(s(x28))))) with σ = [x28 / x7, x27 / x5] which results in the following new constraint:
  

  (5)    (IF(false, s(x26), s(s(x5)), s(x25), s(s(x7)))≥LOOP(s(x26), s(s(x5)), s(plus(x25, s(plus(x25, times(x5, s(x25)))))), s(s(s(x7)))) ⇒ IF(false, s(s(x26)), s(s(x5)), s(s(x25)), s(s(x7)))≥LOOP(s(s(x26)), s(s(x5)), s(plus(s(x25), s(plus(s(x25), times(x5, s(s(x25))))))), s(s(s(x7)))))

  
  
  We simplified constraint (4) using rules (I), (II) which results in the following new constraint:
  

  (6)    (IF(false, s(s(x29)), s(s(x5)), s(0), s(s(x7)))≥LOOP(s(s(x29)), s(s(x5)), s(plus(0, s(plus(0, times(x5, s(0)))))), s(s(s(x7)))))

  
  
  

For Pair LOOP(s(z0), s(s(z1)), s(y_2), s(s(z3))) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(s(z3))) the following chains were created:

• We consider the chain IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3))), LOOP(s(z0), s(s(z1)), s(y_2), s(s(z3))) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(s(z3))) which results in the following constraint:
  

  (7)    (LOOP(s(x12), s(s(x13)), s(plus(x14, s(plus(x14, times(x13, s(x14)))))), s(s(x15)))=LOOP(s(x16), s(s(x17)), s(x18), s(s(x19))) ⇒ LOOP(s(x16), s(s(x17)), s(x18), s(s(x19)))≥IF(le(x16, x18), s(x16), s(s(x17)), s(x18), s(s(x19))))

  
  
  We simplified constraint (7) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint:
  

  (8)    (LOOP(s(x12), s(s(x13)), s(x18), s(s(x15)))≥IF(le(x12, x18), s(x12), s(s(x13)), s(x18), s(s(x15))))

  
  
  

To summarize, we get the following constraints P_≥ for the following pairs.

• IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3)))
  
  • (IF(false, s(x26), s(s(x5)), s(x25), s(s(x7)))≥LOOP(s(x26), s(s(x5)), s(plus(x25, s(plus(x25, times(x5, s(x25)))))), s(s(s(x7)))) ⇒ IF(false, s(s(x26)), s(s(x5)), s(s(x25)), s(s(x7)))≥LOOP(s(s(x26)), s(s(x5)), s(plus(s(x25), s(plus(s(x25), times(x5, s(s(x25))))))), s(s(s(x7)))))
    
  • (IF(false, s(s(x29)), s(s(x5)), s(0), s(s(x7)))≥LOOP(s(s(x29)), s(s(x5)), s(plus(0, s(plus(0, times(x5, s(0)))))), s(s(s(x7)))))
    
  
  
• LOOP(s(z0), s(s(z1)), s(y_2), s(s(z3))) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(s(z3)))
  
  • (LOOP(s(x12), s(s(x13)), s(x18), s(s(x15)))≥IF(le(x12, x18), s(x12), s(s(x13)), s(x18), s(s(x15))))
    
  
  

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(IF(x₁, x₂, x₃, x₄, x₅)) = -1 + x₂ - x₄   
POL(LOOP(x₁, x₂, x₃, x₄)) = -1 + x₁ - x₃   
POL(c) = -1   
POL(false) = 1   
POL(le(x₁, x₂)) = 0   
POL(plus(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 1 + x₁   
POL(times(x₁, x₂)) = 0   
POL(true) = 0   

The following pairs are in P_>:

IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3)))

The following pairs are in P_bound:

IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3)))

The following rules are usable:

0 → times(0, y)
plus(y, times(x, y)) → times(s(x), y)
y → plus(0, y)
s(plus(x, y)) → plus(s(x), y)

(58) Obligation:

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

LOOP(s(z0), s(s(z1)), s(y_2), s(s(z3))) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(s(z3)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)

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

(59) DependencyGraphProof (EQUIVALENT transformation)

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

(60) TRUE