Termination Proof using AProVE

Term Rewriting System R:
[x, y]
even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

EVEN(s(s(x))) -> EVEN(x)
HALF(s(s(x))) -> HALF(x)
PLUS(s(x), y) -> PLUS(x, y)
TIMES(s(x), y) -> IF_TIMES(even(s(x)), s(x), y)
TIMES(s(x), y) -> EVEN(s(x))
IF_TIMES(true, s(x), y) -> PLUS(times(half(s(x)), y), times(half(s(x)), y))
IF_TIMES(true, s(x), y) -> TIMES(half(s(x)), y)
IF_TIMES(true, s(x), y) -> HALF(s(x))
IF_TIMES(false, s(x), y) -> PLUS(y, times(x, y))
IF_TIMES(false, s(x), y) -> TIMES(x, y)

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

Dependency Pair:

EVEN(s(s(x))) -> EVEN(x)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

EVEN(s(s(x))) -> EVEN(x)

one new Dependency Pair is created:

EVEN(s(s(s(s(x''))))) -> EVEN(s(s(x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

Dependency Pair:

EVEN(s(s(s(s(x''))))) -> EVEN(s(s(x'')))

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

EVEN(s(s(s(s(x''))))) -> EVEN(s(s(x'')))

one new Dependency Pair is created:

EVEN(s(s(s(s(s(s(x''''))))))) -> EVEN(s(s(s(s(x'''')))))

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

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

Dependency Pair:

EVEN(s(s(s(s(s(s(x''''))))))) -> EVEN(s(s(s(s(x'''')))))

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

EVEN(s(s(s(s(s(s(x''''))))))) -> EVEN(s(s(s(s(x'''')))))

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

EVEN(x₁) -> EVEN(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

Dependency Pair:

HALF(s(s(x))) -> HALF(x)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

HALF(s(s(x))) -> HALF(x)

one new Dependency Pair is created:

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

Dependency Pair:

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))

one new Dependency Pair is created:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳FwdInst

...

               →DP Problem 9

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

Dependency Pair:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

HALF(x₁) -> HALF(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳FwdInst

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 11

             ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 11

             ↳FwdInst

...

               →DP Problem 12

                 ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 11

             ↳FwdInst

...

               →DP Problem 13

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Narrowing Transformation

Dependency Pairs:

IF_TIMES(false, s(x), y) -> TIMES(x, y)
IF_TIMES(true, s(x), y) -> TIMES(half(s(x)), y)
TIMES(s(x), y) -> IF_TIMES(even(s(x)), s(x), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

TIMES(s(x), y) -> IF_TIMES(even(s(x)), s(x), y)

two new Dependency Pairs are created:

TIMES(s(0), y) -> IF_TIMES(false, s(0), y)
TIMES(s(s(x'')), y) -> IF_TIMES(even(x''), s(s(x'')), y)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

           →DP Problem 14

             ↳Narrowing Transformation

Dependency Pairs:

IF_TIMES(true, s(x), y) -> TIMES(half(s(x)), y)
TIMES(s(s(x'')), y) -> IF_TIMES(even(x''), s(s(x'')), y)
TIMES(s(0), y) -> IF_TIMES(false, s(0), y)
IF_TIMES(false, s(x), y) -> TIMES(x, y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

IF_TIMES(true, s(x), y) -> TIMES(half(s(x)), y)

one new Dependency Pair is created:

IF_TIMES(true, s(s(x'')), y) -> TIMES(s(half(x'')), y)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

           →DP Problem 14

             ↳Nar

...

               →DP Problem 15

                 ↳Narrowing Transformation

Dependency Pairs:

IF_TIMES(true, s(s(x'')), y) -> TIMES(s(half(x'')), y)
TIMES(s(0), y) -> IF_TIMES(false, s(0), y)
IF_TIMES(false, s(x), y) -> TIMES(x, y)
TIMES(s(s(x'')), y) -> IF_TIMES(even(x''), s(s(x'')), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

TIMES(s(s(x'')), y) -> IF_TIMES(even(x''), s(s(x'')), y)

three new Dependency Pairs are created:

TIMES(s(s(0)), y) -> IF_TIMES(true, s(s(0)), y)
TIMES(s(s(s(0))), y) -> IF_TIMES(false, s(s(s(0))), y)
TIMES(s(s(s(s(x')))), y) -> IF_TIMES(even(x'), s(s(s(s(x')))), y)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

           →DP Problem 14

             ↳Nar

...

               →DP Problem 16

                 ↳Narrowing Transformation

Dependency Pairs:

TIMES(s(s(s(s(x')))), y) -> IF_TIMES(even(x'), s(s(s(s(x')))), y)
TIMES(s(s(s(0))), y) -> IF_TIMES(false, s(s(s(0))), y)
TIMES(s(s(0)), y) -> IF_TIMES(true, s(s(0)), y)
IF_TIMES(false, s(x), y) -> TIMES(x, y)
TIMES(s(0), y) -> IF_TIMES(false, s(0), y)
IF_TIMES(true, s(s(x'')), y) -> TIMES(s(half(x'')), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

IF_TIMES(true, s(s(x'')), y) -> TIMES(s(half(x'')), y)

two new Dependency Pairs are created:

IF_TIMES(true, s(s(0)), y) -> TIMES(s(0), y)
IF_TIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

           →DP Problem 14

             ↳Nar

...

               →DP Problem 17

                 ↳Instantiation Transformation

Dependency Pairs:

IF_TIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)
TIMES(s(s(s(0))), y) -> IF_TIMES(false, s(s(s(0))), y)
IF_TIMES(true, s(s(0)), y) -> TIMES(s(0), y)
TIMES(s(s(0)), y) -> IF_TIMES(true, s(s(0)), y)
TIMES(s(0), y) -> IF_TIMES(false, s(0), y)
IF_TIMES(false, s(x), y) -> TIMES(x, y)
TIMES(s(s(s(s(x')))), y) -> IF_TIMES(even(x'), s(s(s(s(x')))), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

IF_TIMES(false, s(x), y) -> TIMES(x, y)

three new Dependency Pairs are created:

IF_TIMES(false, s(0), y'') -> TIMES(0, y'')
IF_TIMES(false, s(s(s(0))), y'') -> TIMES(s(s(0)), y'')
IF_TIMES(false, s(s(s(s(x'0')))), y'') -> TIMES(s(s(s(x'0'))), y'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

           →DP Problem 14

             ↳Nar

...

               →DP Problem 18

                 ↳Forward Instantiation Transformation

Dependency Pairs:

IF_TIMES(false, s(s(s(s(x'0')))), y'') -> TIMES(s(s(s(x'0'))), y'')
TIMES(s(s(s(s(x')))), y) -> IF_TIMES(even(x'), s(s(s(s(x')))), y)
IF_TIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

IF_TIMES(false, s(s(s(s(x'0')))), y'') -> TIMES(s(s(s(x'0'))), y'')

one new Dependency Pair is created:

IF_TIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

           →DP Problem 14

             ↳Nar

...

               →DP Problem 19

                 ↳Argument Filtering and Ordering

Dependency Pairs:

IF_TIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')
IF_TIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)
TIMES(s(s(s(s(x')))), y) -> IF_TIMES(even(x'), s(s(s(s(x')))), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IF_TIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')
IF_TIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)

The following usable rules for innermost can be oriented:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

{true, even}
{0, false}
s > half

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

IF_TIMES(x₁, x₂, x₃) -> x₂
s(x₁) -> s(x₁)
TIMES(x₁, x₂) -> x₁
half(x₁) -> half(x₁)
even(x₁) -> even(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

           →DP Problem 14

             ↳Nar

...

               →DP Problem 20

                 ↳Dependency Graph

Dependency Pair:

TIMES(s(s(s(s(x')))), y) -> IF_TIMES(even(x'), s(s(s(s(x')))), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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