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

         ↳Nar

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

         ↳Nar

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

         ↳Nar

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: Polynomial ordering with Polynomial interpretation:

POL(LE(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

         ↳Nar

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

         ↳Nar

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

         ↳Nar

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

         ↳Nar

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

         ↳Nar

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

         ↳Nar

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'''')))

The following usable rules for innermost w.r.t. to the AFS can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IF_MINUS(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(false) = 0

POL(MINUS(x₁, x₂)) = x₁ + x₂

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(le) = 0

resulting in one new DP problem.
Used Argument Filtering System:

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

     ↳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

         ↳Nar

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

         ↳Nar

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

         ↳Nar

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))

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IF_MINUS(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(0) = 0

POL(false) = 0

POL(MINUS(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

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

         ↳Nar

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

         ↳Nar

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

         ↳Nar

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

         ↳Nar

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:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IF_MINUS(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(0) = 0

POL(false) = 0

POL(MINUS(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

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

     ↳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

         ↳Nar

Dependency Pair:

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

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

         ↳Nar

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

         ↳Nar

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

         ↳Nar

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

                 ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳Nar

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

The following dependency pair can be strictly oriented:

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

The following usable rules for innermost w.r.t. to the AFS can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(false) = 0

POL(minus(x₁, x₂)) = x₁ + x₂

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(if_minus(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(le) = 0

resulting in one new DP problem.
Used Argument Filtering System:

QUOT(x₁, x₂) -> QUOT(x₁, x₂)
s(x₁) -> s(x₁)
minus(x₁, x₂) -> minus(x₁, x₂)
if_minus(x₁, x₂, x₃) -> if_minus(x₁, x₂, x₃)
le(x₁, x₂) -> le

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 22

                 ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳Nar

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

The following dependency pair can be strictly oriented:

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

The following usable rules for innermost w.r.t. to the AFS can be oriented:

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))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

QUOT(x₁, x₂) -> QUOT(x₁, x₂)
s(x₁) -> s(x₁)
if_minus(x₁, x₂, x₃) -> x₂
minus(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 20

             ↳Nar

...

               →DP Problem 24

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

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

         ↳Nar

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Narrowing Transformation

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

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

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

two new Dependency Pairs are created:

LOG(s(s(0))) -> LOG(s(0))
LOG(s(s(s(x'')))) -> LOG(s(s(quot(minus(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

         ↳Nar

           →DP Problem 26

             ↳Narrowing Transformation

Dependency Pair:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(minus(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 Narrowing SCC transformation can be performed.
As a result of transforming the rule

LOG(s(s(s(x'')))) -> LOG(s(s(quot(minus(x'', s(0)), s(s(0))))))

two new Dependency Pairs are created:

LOG(s(s(s(0)))) -> LOG(s(s(quot(0, s(s(0))))))
LOG(s(s(s(s(x'))))) -> LOG(s(s(quot(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 4

         ↳Nar

           →DP Problem 26

             ↳Nar

...

               →DP Problem 27

                 ↳Rewriting Transformation

Dependency Pairs:

LOG(s(s(s(s(x'))))) -> LOG(s(s(quot(if_minus(le(s(x'), s(0)), s(x'), s(0)), s(s(0))))))
LOG(s(s(s(0)))) -> LOG(s(s(quot(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

LOG(s(s(s(0)))) -> LOG(s(s(quot(0, s(s(0))))))

one new Dependency Pair is created:

LOG(s(s(s(0)))) -> LOG(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

         ↳Nar

           →DP Problem 26

             ↳Nar

...

               →DP Problem 28

                 ↳Rewriting Transformation

Dependency Pair:

LOG(s(s(s(s(x'))))) -> LOG(s(s(quot(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

LOG(s(s(s(s(x'))))) -> LOG(s(s(quot(if_minus(le(s(x'), s(0)), s(x'), s(0)), s(s(0))))))

one new Dependency Pair is created:

LOG(s(s(s(s(x'))))) -> LOG(s(s(quot(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

         ↳Nar

           →DP Problem 26

             ↳Nar

...

               →DP Problem 29

                 ↳Argument Filtering and Ordering

Dependency Pair:

LOG(s(s(s(s(x'))))) -> LOG(s(s(quot(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

The following dependency pair can be strictly oriented:

LOG(s(s(s(s(x'))))) -> LOG(s(s(quot(if_minus(le(x', 0), s(x'), s(0)), s(s(0))))))

The following usable rules for innermost w.r.t. to the AFS can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(s(x₁)) = 1 + x₁

POL(LOG(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

LOG(x₁) -> LOG(x₁)
s(x₁) -> s(x₁)
quot(x₁, x₂) -> x₁
if_minus(x₁, x₂, x₃) -> x₂
minus(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Nar

           →DP Problem 26

             ↳Nar

...

               →DP Problem 30

                 ↳Dependency Graph

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.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes

POL(LE(x₁, x₂))	= 1 + x₁ + x₂
POL(s(x₁))	= 1 + x₁

POL(IF_MINUS(x₁, x₂, x₃))	= x₁ + x₂ + x₃
POL(false)	= 0
POL(MINUS(x₁, x₂))	= x₁ + x₂
POL(true)	= 0
POL(s(x₁))	= 1 + x₁
POL(le)	= 0

POL(IF_MINUS(x₁, x₂, x₃))	= x₁ + x₂ + x₃
POL(0)	= 0
POL(false)	= 0
POL(MINUS(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= 1 + x₁

POL(QUOT(x₁, x₂))	= 1 + x₁ + x₂
POL(0)	= 0
POL(false)	= 0
POL(minus(x₁, x₂))	= x₁ + x₂
POL(true)	= 0
POL(s(x₁))	= 1 + x₁
POL(if_minus(x₁, x₂, x₃))	= x₁ + x₂ + x₃
POL(le)	= 0