Termination Proof using AProVE

Term Rewriting System R:
[y, x]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
MINUS(s(x), y) -> IF_MINUS(le(s(x), y), s(x), y)
MINUS(s(x), y) -> LE(s(x), y)
IF_MINUS(false, s(x), y) -> MINUS(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(x, y)
LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))
LOG(s(s(x))) -> QUOT(x, s(s(0)))

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

LE(s(x), s(y)) -> LE(x, y)

one new Dependency Pair is created:

LE(s(s(x'')), s(s(y''))) -> LE(s(x''), s(y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

LE(s(s(x'')), s(s(y''))) -> LE(s(x''), s(y''))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

LE(s(s(x'')), s(s(y''))) -> LE(s(x''), s(y''))

one new Dependency Pair is created:

LE(s(s(s(x''''))), s(s(s(y'''')))) -> LE(s(s(x'''')), s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 6

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

LE(s(s(s(x''''))), s(s(s(y'''')))) -> LE(s(s(x'''')), s(s(y'''')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LE(s(s(s(x''''))), s(s(s(y'''')))) -> LE(s(s(x'''')), s(s(y'''')))

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

LE(x₁, x₂) -> LE(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Narrowing Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

IF_MINUS(false, s(x), y) -> MINUS(x, y)
MINUS(s(x), y) -> IF_MINUS(le(s(x), y), s(x), y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MINUS(s(x), y) -> IF_MINUS(le(s(x), y), s(x), y)

two new Dependency Pairs are created:

MINUS(s(x''), 0) -> IF_MINUS(false, s(x''), 0)
MINUS(s(x''), s(y'')) -> IF_MINUS(le(x'', y''), s(x''), s(y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Narrowing Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

MINUS(s(x''), s(y'')) -> IF_MINUS(le(x'', y''), s(x''), s(y''))
MINUS(s(x''), 0) -> IF_MINUS(false, s(x''), 0)
IF_MINUS(false, s(x), y) -> MINUS(x, y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MINUS(s(x''), s(y'')) -> IF_MINUS(le(x'', y''), s(x''), s(y''))

three new Dependency Pairs are created:

MINUS(s(0), s(y''')) -> IF_MINUS(true, s(0), s(y'''))
MINUS(s(s(x')), s(0)) -> IF_MINUS(false, s(s(x')), s(0))
MINUS(s(s(x')), s(s(y'))) -> IF_MINUS(le(x', y'), s(s(x')), s(s(y')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 9

                 ↳Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

MINUS(s(s(x')), s(s(y'))) -> IF_MINUS(le(x', y'), s(s(x')), s(s(y')))
MINUS(s(s(x')), s(0)) -> IF_MINUS(false, s(s(x')), s(0))
IF_MINUS(false, s(x), y) -> MINUS(x, y)
MINUS(s(x''), 0) -> IF_MINUS(false, s(x''), 0)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_MINUS(false, s(x), y) -> MINUS(x, y)

three new Dependency Pairs are created:

IF_MINUS(false, s(x'), 0) -> MINUS(x', 0)
IF_MINUS(false, s(s(x''')), s(0)) -> MINUS(s(x'''), s(0))
IF_MINUS(false, s(s(x'0')), s(s(y'''))) -> MINUS(s(x'0'), s(s(y''')))

The transformation is resulting in three new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 10

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

IF_MINUS(false, s(s(x'0')), s(s(y'''))) -> MINUS(s(x'0'), s(s(y''')))
MINUS(s(s(x')), s(s(y'))) -> IF_MINUS(le(x', y'), s(s(x')), s(s(y')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_MINUS(false, s(s(x'0')), s(s(y'''))) -> MINUS(s(x'0'), s(s(y''')))

one new Dependency Pair is created:

IF_MINUS(false, s(s(s(x'''))), s(s(y''''))) -> MINUS(s(s(x''')), s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 13

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

IF_MINUS(false, s(s(s(x'''))), s(s(y''''))) -> MINUS(s(s(x''')), s(s(y'''')))
MINUS(s(s(x')), s(s(y'))) -> IF_MINUS(le(x', y'), s(s(x')), s(s(y')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_MINUS(false, s(s(s(x'''))), s(s(y''''))) -> MINUS(s(s(x''')), s(s(y'''')))

IF_MINUS(x₁, x₂, x₃) -> x₂
s(x₁) -> s(x₁)
MINUS(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 17

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

MINUS(s(s(x')), s(s(y'))) -> IF_MINUS(le(x', y'), s(s(x')), s(s(y')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 11

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

IF_MINUS(false, s(s(x''')), s(0)) -> MINUS(s(x'''), s(0))
MINUS(s(s(x')), s(0)) -> IF_MINUS(false, s(s(x')), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_MINUS(false, s(s(x''')), s(0)) -> MINUS(s(x'''), s(0))

one new Dependency Pair is created:

IF_MINUS(false, s(s(s(x''''))), s(0)) -> MINUS(s(s(x'''')), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 14

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

IF_MINUS(false, s(s(s(x''''))), s(0)) -> MINUS(s(s(x'''')), s(0))
MINUS(s(s(x')), s(0)) -> IF_MINUS(false, s(s(x')), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_MINUS(false, s(s(s(x''''))), s(0)) -> MINUS(s(s(x'''')), s(0))

IF_MINUS(x₁, x₂, x₃) -> x₂
s(x₁) -> s(x₁)
MINUS(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 18

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

MINUS(s(s(x')), s(0)) -> IF_MINUS(false, s(s(x')), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 12

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

IF_MINUS(false, s(x'), 0) -> MINUS(x', 0)
MINUS(s(x''), 0) -> IF_MINUS(false, s(x''), 0)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_MINUS(false, s(x'), 0) -> MINUS(x', 0)

one new Dependency Pair is created:

IF_MINUS(false, s(s(x'''')), 0) -> MINUS(s(x''''), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 15

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

IF_MINUS(false, s(s(x'''')), 0) -> MINUS(s(x''''), 0)
MINUS(s(x''), 0) -> IF_MINUS(false, s(x''), 0)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

MINUS(s(x''), 0) -> IF_MINUS(false, s(x''), 0)

one new Dependency Pair is created:

MINUS(s(s(x'''''')), 0) -> IF_MINUS(false, s(s(x'''''')), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 16

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

MINUS(s(s(x'''''')), 0) -> IF_MINUS(false, s(s(x'''''')), 0)
IF_MINUS(false, s(s(x'''')), 0) -> MINUS(s(x''''), 0)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_MINUS(false, s(s(x'''')), 0) -> MINUS(s(x''''), 0)

MINUS(x₁, x₂) -> x₁
s(x₁) -> s(x₁)
IF_MINUS(x₁, x₂, x₃) -> x₂

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 19

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

MINUS(s(s(x'''''')), 0) -> IF_MINUS(false, s(s(x'''''')), 0)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Narrowing Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

two new Dependency Pairs are created:

QUOT(s(0), s(y'')) -> QUOT(0, s(y''))
QUOT(s(s(x'')), s(y'')) -> QUOT(if_minus(le(s(x''), y''), s(x''), y''), s(y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Narrowing Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(x'')), s(y'')) -> QUOT(if_minus(le(s(x''), y''), s(x''), y''), s(y''))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(x'')), s(y'')) -> QUOT(if_minus(le(s(x''), y''), s(x''), y''), s(y''))

two new Dependency Pairs are created:

QUOT(s(s(x''')), s(0)) -> QUOT(if_minus(false, s(x'''), 0), s(0))
QUOT(s(s(x''')), s(s(y'))) -> QUOT(if_minus(le(x''', y'), s(x'''), s(y')), s(s(y')))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 21

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(x''')), s(0)) -> QUOT(if_minus(false, s(x'''), 0), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(x''')), s(0)) -> QUOT(if_minus(false, s(x'''), 0), s(0))

one new Dependency Pair is created:

QUOT(s(s(x''')), s(0)) -> QUOT(s(minus(x''', 0)), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 23

                 ↳Narrowing Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(x''')), s(0)) -> QUOT(s(minus(x''', 0)), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(x''')), s(0)) -> QUOT(s(minus(x''', 0)), s(0))

two new Dependency Pairs are created:

QUOT(s(s(0)), s(0)) -> QUOT(s(0), s(0))
QUOT(s(s(s(x'))), s(0)) -> QUOT(s(if_minus(le(s(x'), 0), s(x'), 0)), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 25

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(x'))), s(0)) -> QUOT(s(if_minus(le(s(x'), 0), s(x'), 0)), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(x'))), s(0)) -> QUOT(s(if_minus(le(s(x'), 0), s(x'), 0)), s(0))

one new Dependency Pair is created:

QUOT(s(s(s(x'))), s(0)) -> QUOT(s(if_minus(false, s(x'), 0)), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 28

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(x'))), s(0)) -> QUOT(s(if_minus(false, s(x'), 0)), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(x'))), s(0)) -> QUOT(s(if_minus(false, s(x'), 0)), s(0))

one new Dependency Pair is created:

QUOT(s(s(s(x'))), s(0)) -> QUOT(s(s(minus(x', 0))), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(s(s(x'))), s(0)) -> QUOT(s(s(minus(x', 0))), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(if_minus(le(x', 0), s(x'), s(0))), s(s(0)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y'')))))) -> QUOT(if_minus(le(x', y''), s(s(s(s(x')))), s(s(s(s(y''))))), s(s(s(s(s(y''))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(x'', 0), s(s(x'')), s(s(0)))), s(s(s(0))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(x', 0), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 22

                 ↳Narrowing Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(x''')), s(s(y'))) -> QUOT(if_minus(le(x''', y'), s(x'''), s(y')), s(s(y')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(x''')), s(s(y'))) -> QUOT(if_minus(le(x''', y'), s(x'''), s(y')), s(s(y')))

three new Dependency Pairs are created:

QUOT(s(s(0)), s(s(y''))) -> QUOT(if_minus(true, s(0), s(y'')), s(s(y'')))
QUOT(s(s(s(x'))), s(s(0))) -> QUOT(if_minus(false, s(s(x')), s(0)), s(s(0)))
QUOT(s(s(s(x'))), s(s(s(y'')))) -> QUOT(if_minus(le(x', y''), s(s(x')), s(s(y''))), s(s(s(y''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 24

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(x'))), s(s(s(y'')))) -> QUOT(if_minus(le(x', y''), s(s(x')), s(s(y''))), s(s(s(y''))))
QUOT(s(s(s(x'))), s(s(0))) -> QUOT(if_minus(false, s(s(x')), s(0)), s(s(0)))
QUOT(s(s(0)), s(s(y''))) -> QUOT(if_minus(true, s(0), s(y'')), s(s(y'')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(0)), s(s(y''))) -> QUOT(if_minus(true, s(0), s(y'')), s(s(y'')))

one new Dependency Pair is created:

QUOT(s(s(0)), s(s(y''))) -> QUOT(0, s(s(y'')))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 26

                 ↳Narrowing Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(x'))), s(s(s(y'')))) -> QUOT(if_minus(le(x', y''), s(s(x')), s(s(y''))), s(s(s(y''))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(x'))), s(s(s(y'')))) -> QUOT(if_minus(le(x', y''), s(s(x')), s(s(y''))), s(s(s(y''))))

three new Dependency Pairs are created:

QUOT(s(s(s(0))), s(s(s(y''')))) -> QUOT(if_minus(true, s(s(0)), s(s(y'''))), s(s(s(y'''))))
QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(if_minus(false, s(s(s(x''))), s(s(0))), s(s(s(0))))
QUOT(s(s(s(s(x'')))), s(s(s(s(y'))))) -> QUOT(if_minus(le(x'', y'), s(s(s(x''))), s(s(s(y')))), s(s(s(s(y')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 29

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(s(x'')))), s(s(s(s(y'))))) -> QUOT(if_minus(le(x'', y'), s(s(s(x''))), s(s(s(y')))), s(s(s(s(y')))))
QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(if_minus(false, s(s(s(x''))), s(s(0))), s(s(s(0))))
QUOT(s(s(s(0))), s(s(s(y''')))) -> QUOT(if_minus(true, s(s(0)), s(s(y'''))), s(s(s(y'''))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(0))), s(s(s(y''')))) -> QUOT(if_minus(true, s(s(0)), s(s(y'''))), s(s(s(y'''))))

one new Dependency Pair is created:

QUOT(s(s(s(0))), s(s(s(y''')))) -> QUOT(0, s(s(s(y'''))))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 32

                 ↳Narrowing Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(s(x'')))), s(s(s(s(y'))))) -> QUOT(if_minus(le(x'', y'), s(s(s(x''))), s(s(s(y')))), s(s(s(s(y')))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(s(x'')))), s(s(s(s(y'))))) -> QUOT(if_minus(le(x'', y'), s(s(s(x''))), s(s(s(y')))), s(s(s(s(y')))))

three new Dependency Pairs are created:

QUOT(s(s(s(s(0)))), s(s(s(s(y''))))) -> QUOT(if_minus(true, s(s(s(0))), s(s(s(y'')))), s(s(s(s(y'')))))
QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(if_minus(false, s(s(s(s(x')))), s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y'')))))) -> QUOT(if_minus(le(x', y''), s(s(s(s(x')))), s(s(s(s(y''))))), s(s(s(s(s(y''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 35

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y'')))))) -> QUOT(if_minus(le(x', y''), s(s(s(s(x')))), s(s(s(s(y''))))), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(if_minus(false, s(s(s(s(x')))), s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(s(s(s(0)))), s(s(s(s(y''))))) -> QUOT(if_minus(true, s(s(s(0))), s(s(s(y'')))), s(s(s(s(y'')))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(s(0)))), s(s(s(s(y''))))) -> QUOT(if_minus(true, s(s(s(0))), s(s(s(y'')))), s(s(s(s(y'')))))

one new Dependency Pair is created:

QUOT(s(s(s(s(0)))), s(s(s(s(y''))))) -> QUOT(0, s(s(s(s(y'')))))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(s(s(x'))), s(0)) -> QUOT(s(s(minus(x', 0))), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(if_minus(le(x', 0), s(x'), s(0))), s(s(0)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y'')))))) -> QUOT(if_minus(le(x', y''), s(s(s(s(x')))), s(s(s(s(y''))))), s(s(s(s(s(y''))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(x'', 0), s(s(x'')), s(s(0)))), s(s(s(0))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(x', 0), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 39

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(if_minus(false, s(s(s(s(x')))), s(s(s(0)))), s(s(s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(if_minus(false, s(s(s(s(x')))), s(s(s(0)))), s(s(s(s(0)))))

one new Dependency Pair is created:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(minus(s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 41

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(minus(s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(minus(s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

one new Dependency Pair is created:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(s(s(s(x'))), s(s(s(0)))), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 43

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(s(s(s(x'))), s(s(s(0)))), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(s(s(s(x'))), s(s(s(0)))), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

one new Dependency Pair is created:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(s(s(x')), s(s(0))), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 45

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(s(s(x')), s(s(0))), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(s(s(x')), s(s(0))), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

one new Dependency Pair is created:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(s(x'), s(0)), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 46

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(s(x'), s(0)), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(s(x'), s(0)), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

one new Dependency Pair is created:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(x', 0), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(s(s(x'))), s(0)) -> QUOT(s(s(minus(x', 0))), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(if_minus(le(x', 0), s(x'), s(0))), s(s(0)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y'')))))) -> QUOT(if_minus(le(x', y''), s(s(s(s(x')))), s(s(s(s(y''))))), s(s(s(s(s(y''))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(x'', 0), s(s(x'')), s(s(0)))), s(s(s(0))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(x', 0), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 33

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(if_minus(false, s(s(s(x''))), s(s(0))), s(s(s(0))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(if_minus(false, s(s(s(x''))), s(s(0))), s(s(s(0))))

one new Dependency Pair is created:

QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(minus(s(s(x'')), s(s(0)))), s(s(s(0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 36

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(minus(s(s(x'')), s(s(0)))), s(s(s(0))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(minus(s(s(x'')), s(s(0)))), s(s(s(0))))

one new Dependency Pair is created:

QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(s(s(x'')), s(s(0))), s(s(x'')), s(s(0)))), s(s(s(0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 40

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(s(s(x'')), s(s(0))), s(s(x'')), s(s(0)))), s(s(s(0))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(s(s(x'')), s(s(0))), s(s(x'')), s(s(0)))), s(s(s(0))))

one new Dependency Pair is created:

QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(s(x''), s(0)), s(s(x'')), s(s(0)))), s(s(s(0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 42

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(s(x''), s(0)), s(s(x'')), s(s(0)))), s(s(s(0))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(s(x''), s(0)), s(s(x'')), s(s(0)))), s(s(s(0))))

one new Dependency Pair is created:

QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(x'', 0), s(s(x'')), s(s(0)))), s(s(s(0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(s(s(x'))), s(0)) -> QUOT(s(s(minus(x', 0))), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(if_minus(le(x', 0), s(x'), s(0))), s(s(0)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y'')))))) -> QUOT(if_minus(le(x', y''), s(s(s(s(x')))), s(s(s(s(y''))))), s(s(s(s(s(y''))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(x'', 0), s(s(x'')), s(s(0)))), s(s(s(0))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(x', 0), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 27

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(x'))), s(s(0))) -> QUOT(if_minus(false, s(s(x')), s(0)), s(s(0)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(x'))), s(s(0))) -> QUOT(if_minus(false, s(s(x')), s(0)), s(s(0)))

one new Dependency Pair is created:

QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(minus(s(x'), s(0))), s(s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 30

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(minus(s(x'), s(0))), s(s(0)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(minus(s(x'), s(0))), s(s(0)))

one new Dependency Pair is created:

QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(if_minus(le(s(x'), s(0)), s(x'), s(0))), s(s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 34

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(if_minus(le(s(x'), s(0)), s(x'), s(0))), s(s(0)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(if_minus(le(s(x'), s(0)), s(x'), s(0))), s(s(0)))

one new Dependency Pair is created:

QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(if_minus(le(x', 0), s(x'), s(0))), s(s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(s(s(x'))), s(0)) -> QUOT(s(s(minus(x', 0))), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(if_minus(le(x', 0), s(x'), s(0))), s(s(0)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y'')))))) -> QUOT(if_minus(le(x', y''), s(s(s(s(x')))), s(s(s(s(y''))))), s(s(s(s(s(y''))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(x'', 0), s(s(x'')), s(s(0)))), s(s(s(0))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(x', 0), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(s(s(x'))), s(0)) -> QUOT(s(s(minus(x', 0))), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(x'))), s(s(0))) -> QUOT(s(if_minus(le(x', 0), s(x'), s(0))), s(s(0)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y'')))))) -> QUOT(if_minus(le(x', y''), s(s(s(s(x')))), s(s(s(s(y''))))), s(s(s(s(s(y''))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(x'')))), s(s(s(0)))) -> QUOT(s(if_minus(le(x'', 0), s(s(x'')), s(s(0)))), s(s(s(0))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(x'))))), s(s(s(s(0))))) -> QUOT(s(if_minus(le(x', 0), s(s(s(x'))), s(s(s(0))))), s(s(s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost
Dependency Pair:
LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Strategy:
innermost

Innermost Termination of R could not be shown.
Duration:
0:02 minutes