Termination Proof using AProVE

Term Rewriting System R:
[Y, X, X1, X2, X3]
a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_MINUS}(s(X), s(Y)) -> A_{_MINUS}(X, Y)
A_{_GEQ}(s(X), s(Y)) -> A_{_GEQ}(X, Y)
A_{_DIV}(s(X), s(Y)) -> A_{_IF}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
A_{_DIV}(s(X), s(Y)) -> A_{_GEQ}(X, Y)
A_{_IF}(true, X, Y) -> MARK(X)
A_{_IF}(false, X, Y) -> MARK(Y)
MARK(minus(X1, X2)) -> A_{_MINUS}(X1, X2)
MARK(geq(X1, X2)) -> A_{_GEQ}(X1, X2)
MARK(div(X1, X2)) -> A_{_DIV}(mark(X1), X2)
MARK(div(X1, X2)) -> MARK(X1)
MARK(if(X1, X2, X3)) -> A_{_IF}(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

Dependency Pair:

A_{_MINUS}(s(X), s(Y)) -> A_{_MINUS}(X, Y)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Strategy:

innermost

The following dependency pair can be strictly oriented:

A_{_MINUS}(s(X), s(Y)) -> A_{_MINUS}(X, Y)

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

A_{_MINUS}(x₁, x₂) -> A_{_MINUS}(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

A_{_GEQ}(s(X), s(Y)) -> A_{_GEQ}(X, Y)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Strategy:

innermost

The following dependency pair can be strictly oriented:

A_{_GEQ}(s(X), s(Y)) -> A_{_GEQ}(X, Y)

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

A_{_GEQ}(x₁, x₂) -> A_{_GEQ}(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(if(X1, X2, X3)) -> MARK(X1)
A_{_IF}(false, X, Y) -> MARK(Y)
MARK(if(X1, X2, X3)) -> A_{_IF}(mark(X1), X2, X3)
MARK(div(X1, X2)) -> MARK(X1)
MARK(div(X1, X2)) -> A_{_DIV}(mark(X1), X2)
A_{_IF}(true, X, Y) -> MARK(X)
A_{_DIV}(s(X), s(Y)) -> A_{_IF}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Strategy:

innermost

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

A_{_DIV}(s(X), s(Y)) -> A_{_IF}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)

four new Dependency Pairs are created:

A_{_DIV}(s(X''), s(0)) -> A_{_IF}(true, s(div(minus(X'', 0), s(0))), 0)
A_{_DIV}(s(0), s(s(Y''))) -> A_{_IF}(false, s(div(minus(0, s(Y'')), s(s(Y'')))), 0)
A_{_DIV}(s(s(X'')), s(s(Y''))) -> A_{_IF}(a_{_geq}(X'', Y''), s(div(minus(s(X''), s(Y'')), s(s(Y'')))), 0)
A_{_DIV}(s(X'), s(Y')) -> A_{_IF}(geq(X', Y'), s(div(minus(X', Y'), s(Y'))), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pairs:

A_{_DIV}(s(s(X'')), s(s(Y''))) -> A_{_IF}(a_{_geq}(X'', Y''), s(div(minus(s(X''), s(Y'')), s(s(Y'')))), 0)
A_{_DIV}(s(0), s(s(Y''))) -> A_{_IF}(false, s(div(minus(0, s(Y'')), s(s(Y'')))), 0)
MARK(if(X1, X2, X3)) -> MARK(X1)
A_{_IF}(false, X, Y) -> MARK(Y)
MARK(if(X1, X2, X3)) -> A_{_IF}(mark(X1), X2, X3)
MARK(div(X1, X2)) -> MARK(X1)
A_{_IF}(true, X, Y) -> MARK(X)
A_{_DIV}(s(X''), s(0)) -> A_{_IF}(true, s(div(minus(X'', 0), s(0))), 0)
MARK(div(X1, X2)) -> A_{_DIV}(mark(X1), X2)
MARK(s(X)) -> MARK(X)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Strategy:

innermost

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

MARK(div(X1, X2)) -> A_{_DIV}(mark(X1), X2)

eight new Dependency Pairs are created:

MARK(div(minus(X1'', X2''), X2)) -> A_{_DIV}(a_{_minus}(X1'', X2''), X2)
MARK(div(geq(X1'', X2''), X2)) -> A_{_DIV}(a_{_geq}(X1'', X2''), X2)
MARK(div(div(X1'', X2''), X2)) -> A_{_DIV}(a_{_div}(mark(X1''), X2''), X2)
MARK(div(if(X1'', X2'', X3'), X2)) -> A_{_DIV}(a_{_if}(mark(X1''), X2'', X3'), X2)
MARK(div(0, X2)) -> A_{_DIV}(0, X2)
MARK(div(s(X'), X2)) -> A_{_DIV}(s(mark(X')), X2)
MARK(div(true, X2)) -> A_{_DIV}(true, X2)
MARK(div(false, X2)) -> A_{_DIV}(false, X2)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

MARK(div(s(X'), X2)) -> A_{_DIV}(s(mark(X')), X2)
MARK(div(if(X1'', X2'', X3'), X2)) -> A_{_DIV}(a_{_if}(mark(X1''), X2'', X3'), X2)
MARK(div(div(X1'', X2''), X2)) -> A_{_DIV}(a_{_div}(mark(X1''), X2''), X2)
MARK(div(geq(X1'', X2''), X2)) -> A_{_DIV}(a_{_geq}(X1'', X2''), X2)
A_{_DIV}(s(0), s(s(Y''))) -> A_{_IF}(false, s(div(minus(0, s(Y'')), s(s(Y'')))), 0)
A_{_DIV}(s(X''), s(0)) -> A_{_IF}(true, s(div(minus(X'', 0), s(0))), 0)
MARK(div(minus(X1'', X2''), X2)) -> A_{_DIV}(a_{_minus}(X1'', X2''), X2)
MARK(s(X)) -> MARK(X)
MARK(if(X1, X2, X3)) -> MARK(X1)
A_{_IF}(false, X, Y) -> MARK(Y)
MARK(if(X1, X2, X3)) -> A_{_IF}(mark(X1), X2, X3)
MARK(div(X1, X2)) -> MARK(X1)
A_{_IF}(true, X, Y) -> MARK(X)
A_{_DIV}(s(s(X'')), s(s(Y''))) -> A_{_IF}(a_{_geq}(X'', Y''), s(div(minus(s(X''), s(Y'')), s(s(Y'')))), 0)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Strategy:

innermost

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

MARK(if(X1, X2, X3)) -> A_{_IF}(mark(X1), X2, X3)

eight new Dependency Pairs are created:

MARK(if(minus(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_minus}(X1'', X2''), X2, X3)
MARK(if(geq(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_geq}(X1'', X2''), X2, X3)
MARK(if(div(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_div}(mark(X1''), X2''), X2, X3)
MARK(if(if(X1'', X2'', X3''), X2, X3)) -> A_{_IF}(a_{_if}(mark(X1''), X2'', X3''), X2, X3)
MARK(if(0, X2, X3)) -> A_{_IF}(0, X2, X3)
MARK(if(s(X'), X2, X3)) -> A_{_IF}(s(mark(X')), X2, X3)
MARK(if(true, X2, X3)) -> A_{_IF}(true, X2, X3)
MARK(if(false, X2, X3)) -> A_{_IF}(false, X2, X3)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

MARK(if(false, X2, X3)) -> A_{_IF}(false, X2, X3)
MARK(if(true, X2, X3)) -> A_{_IF}(true, X2, X3)
MARK(if(if(X1'', X2'', X3''), X2, X3)) -> A_{_IF}(a_{_if}(mark(X1''), X2'', X3''), X2, X3)
MARK(if(div(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_div}(mark(X1''), X2''), X2, X3)
MARK(if(geq(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_geq}(X1'', X2''), X2, X3)
MARK(if(minus(X1'', X2''), X2, X3)) -> A_{_IF}(a_{_minus}(X1'', X2''), X2, X3)
MARK(div(if(X1'', X2'', X3'), X2)) -> A_{_DIV}(a_{_if}(mark(X1''), X2'', X3'), X2)
MARK(div(div(X1'', X2''), X2)) -> A_{_DIV}(a_{_div}(mark(X1''), X2''), X2)
A_{_DIV}(s(s(X'')), s(s(Y''))) -> A_{_IF}(a_{_geq}(X'', Y''), s(div(minus(s(X''), s(Y'')), s(s(Y'')))), 0)
MARK(div(geq(X1'', X2''), X2)) -> A_{_DIV}(a_{_geq}(X1'', X2''), X2)
A_{_IF}(false, X, Y) -> MARK(Y)
A_{_DIV}(s(0), s(s(Y''))) -> A_{_IF}(false, s(div(minus(0, s(Y'')), s(s(Y'')))), 0)
MARK(div(minus(X1'', X2''), X2)) -> A_{_DIV}(a_{_minus}(X1'', X2''), X2)
MARK(s(X)) -> MARK(X)
MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(div(X1, X2)) -> MARK(X1)
A_{_IF}(true, X, Y) -> MARK(X)
A_{_DIV}(s(X''), s(0)) -> A_{_IF}(true, s(div(minus(X'', 0), s(0))), 0)
MARK(div(s(X'), X2)) -> A_{_DIV}(s(mark(X')), X2)

Rules:

a_{_minus}(0, Y) -> 0
a_{_minus}(s(X), s(Y)) -> a_{_minus}(X, Y)
a_{_minus}(X1, X2) -> minus(X1, X2)
a_{_geq}(X, 0) -> true
a_{_geq}(0, s(Y)) -> false
a_{_geq}(s(X), s(Y)) -> a_{_geq}(X, Y)
a_{_geq}(X1, X2) -> geq(X1, X2)
a_{_div}(0, s(Y)) -> 0
a_{_div}(s(X), s(Y)) -> a_{_if}(a_{_geq}(X, Y), s(div(minus(X, Y), s(Y))), 0)
a_{_div}(X1, X2) -> div(X1, X2)
a_{_if}(true, X, Y) -> mark(X)
a_{_if}(false, X, Y) -> mark(Y)
a_{_if}(X1, X2, X3) -> if(X1, X2, X3)
mark(minus(X1, X2)) -> a_{_minus}(X1, X2)
mark(geq(X1, X2)) -> a_{_geq}(X1, X2)
mark(div(X1, X2)) -> a_{_div}(mark(X1), X2)
mark(if(X1, X2, X3)) -> a_{_if}(mark(X1), X2, X3)
mark(0) -> 0
mark(s(X)) -> s(mark(X))
mark(true) -> true
mark(false) -> false

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:15 minutes