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)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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)
INC(s(x)) → INC(x)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) → MINUS(x, y)
LOG(x) → LOG2(x, 0)
LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))
LOG2(x, y) → LE(x, 0)
LOG2(x, y) → LE(x, s(0))
LOG2(x, y) → INC(y)
IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF2(false, x, y) → QUOT(x, s(s(0)))

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)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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 5 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)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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:

QUOT(s(x), s(y)) → QUOT(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)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(0, y) → 0
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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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))
inc(0)
inc(s(x0))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

(18) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

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

POL(0) = 0   
POL(QUOT(x₁, x₂)) = x₁   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented:

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

(20) Obligation:

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

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

The set Q consists of the following terms:

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

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

INC(s(x)) → INC(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)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

(24) 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.

(25) Obligation:

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

INC(s(x)) → INC(x)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

(26) 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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

(27) Obligation:

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

INC(s(x)) → INC(x)

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

(28) 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:

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

(29) TRUE

(30) 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)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

(31) 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.

(32) 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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

(33) 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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

(34) 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.

(35) 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

(36) TRUE

(37) Obligation:

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

LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))
IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), 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)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

(38) 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.

(39) Obligation:

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

LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))
IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

(40) 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)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

(41) Obligation:

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

LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))
IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(42) RemovalProof (SOUND transformation)

In the following pairs the term without variables s(s(0)) is replaced by the fresh variable x_removed.
Pair: IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
Positions in right side of the pair:

• [0,1]

The new variable was added to all pairs as a new argument[CONREM].

(43) Obligation:

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

LOG2(x, y, x_removed) → IF(le(x, 0), le(x, s(0)), x, inc(y), x_removed)
IF(false, b, x, y, x_removed) → IF2(b, x, y, x_removed)
IF2(false, x, y, x_removed) → LOG2(quot(x, x_removed), y, x_removed)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(44) RemovalProof (SOUND transformation)

In the following pairs the term without variables s(s(0)) is replaced by the fresh variable x_removed.
Pair: IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
Positions in right side of the pair:

• [0,1]

The new variable was added to all pairs as a new argument[CONREM].

(45) Obligation:

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

LOG2(x, y, x_removed) → IF(le(x, 0), le(x, s(0)), x, inc(y), x_removed)
IF(false, b, x, y, x_removed) → IF2(b, x, y, x_removed)
IF2(false, x, y, x_removed) → LOG2(quot(x, x_removed), y, x_removed)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(46) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y)) at position [0] we obtained the following new rules [LPAR04]:

LOG2(0, y1) → IF(true, le(0, s(0)), 0, inc(y1))
LOG2(s(x0), y1) → IF(false, le(s(x0), s(0)), s(x0), inc(y1))

(47) Obligation:

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

IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
LOG2(0, y1) → IF(true, le(0, s(0)), 0, inc(y1))
LOG2(s(x0), y1) → IF(false, le(s(x0), s(0)), s(x0), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(48) DependencyGraphProof (EQUIVALENT transformation)

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

(49) Obligation:

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

IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
LOG2(s(x0), y1) → IF(false, le(s(x0), s(0)), s(x0), inc(y1))
IF(false, b, x, y) → IF2(b, x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(50) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOG2(s(x0), y1) → IF(false, le(s(x0), s(0)), s(x0), inc(y1)) at position [1] we obtained the following new rules [LPAR04]:

LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))

(51) Obligation:

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

IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF(false, b, x, y) → IF2(b, x, y)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(52) 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.

(53) Obligation:

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

IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF(false, b, x, y) → IF2(b, x, y)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(54) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF2(false, x, y) → LOG2(quot(x, s(s(0))), y) at position [0] we obtained the following new rules [LPAR04]:

IF2(false, 0, y1) → LOG2(0, y1)
IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)

(55) Obligation:

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

IF(false, b, x, y) → IF2(b, x, y)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF2(false, 0, y1) → LOG2(0, y1)
IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(56) DependencyGraphProof (EQUIVALENT transformation)

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

(57) Obligation:

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF(false, b, x, y) → IF2(b, x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(58) Instantiation (EQUIVALENT transformation)

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

IF(false, y_0, s(z0), y_1) → IF2(y_0, s(z0), y_1)

(59) Obligation:

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF(false, y_0, s(z0), y_1) → IF2(y_0, s(z0), y_1)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(60) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF(false, y_0, s(z0), y_1) → IF2(y_0, s(z0), y_1) we obtained the following new rules [LPAR04]:

IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)

(61) Obligation:

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(62) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1)) at position [1] we obtained the following new rules [LPAR04]:

LOG2(s(0), y1) → IF(false, true, s(0), inc(y1))
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))

(63) Obligation:

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)
LOG2(s(0), y1) → IF(false, true, s(0), inc(y1))
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(64) DependencyGraphProof (EQUIVALENT transformation)

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

(65) Obligation:

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

LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)
IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
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))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(66) 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.

(67) Obligation:

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

LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)
IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, 0) → x
inc(0) → 0
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(68) 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))

(69) Obligation:

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

LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)
IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, 0) → x
inc(0) → 0
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(70) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, false, s(x1), x2) → IF2(false, s(x1), x2) we obtained the following new rules [LPAR04]:

IF(false, false, s(s(z0)), y_0) → IF2(false, s(s(z0)), y_0)

(71) Obligation:

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

LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
IF(false, false, s(s(z0)), y_0) → IF2(false, s(s(z0)), y_0)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, 0) → x
inc(0) → 0
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(72) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1) we obtained the following new rules [LPAR04]:

IF2(false, s(s(z0)), z1) → LOG2(s(quot(minus(s(z0), s(0)), s(s(0)))), z1)

(73) Obligation:

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

LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(s(z0)), y_0) → IF2(false, s(s(z0)), y_0)
IF2(false, s(s(z0)), z1) → LOG2(s(quot(minus(s(z0), s(0)), s(s(0)))), z1)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, 0) → x
inc(0) → 0
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(74) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF2(false, s(s(z0)), z1) → LOG2(s(quot(minus(s(z0), s(0)), s(s(0)))), z1) at position [0,0,0] we obtained the following new rules [LPAR04]:

IF2(false, s(s(z0)), z1) → LOG2(s(quot(minus(z0, 0), s(s(0)))), z1)

(75) Obligation:

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

LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(s(z0)), y_0) → IF2(false, s(s(z0)), y_0)
IF2(false, s(s(z0)), z1) → LOG2(s(quot(minus(z0, 0), s(s(0)))), z1)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, 0) → x
inc(0) → 0
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(76) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF2(false, s(s(z0)), z1) → LOG2(s(quot(minus(z0, 0), s(s(0)))), z1) at position [0,0,0] we obtained the following new rules [LPAR04]:

IF2(false, s(s(z0)), z1) → LOG2(s(quot(z0, s(s(0)))), z1)

(77) Obligation:

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

LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(s(z0)), y_0) → IF2(false, s(s(z0)), y_0)
IF2(false, s(s(z0)), z1) → LOG2(s(quot(z0, s(s(0)))), z1)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, 0) → x
inc(0) → 0
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(78) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(false, false, s(s(z0)), y_0) → IF2(false, s(s(z0)), y_0)

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

POL(0) = 0   
POL(IF(x₁, x₂, x₃, x₄)) = 1 + x₃   
POL(IF2(x₁, x₂, x₃)) = x₂   
POL(LOG2(x₁, x₂)) = 1 + x₁   
POL(false) = 0   
POL(inc(x₁)) = 0   
POL(minus(x₁, x₂)) = x₁   
POL(quot(x₁, x₂)) = x₁   
POL(s(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented:

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

(79) Obligation:

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

LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF2(false, s(s(z0)), z1) → LOG2(s(quot(z0, s(s(0)))), z1)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, 0) → x
inc(0) → 0
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(80) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(81) TRUE