Termination Proof using AProVE

Term Rewriting System R:
[Y, X]
minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(s(X), s(Y)) -> MINUS(X, Y)
GEQ(s(X), s(Y)) -> GEQ(X, Y)
DIV(s(X), s(Y)) -> IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
DIV(s(X), s(Y)) -> GEQ(X, Y)
DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pair:

MINUS(s(X), s(Y)) -> MINUS(X, Y)

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

The following dependency pair can be strictly oriented:

MINUS(s(X), s(Y)) -> MINUS(X, Y)

The following rules can be oriented:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

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

div > geq
div > {true, 0}
div > if
div > {false, s}

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

MINUS(x₁, x₂) -> MINUS(x₁, x₂)
s(x₁) -> s(x₁)
minus(x₁, x₂) -> x₁
geq(x₁, x₂) -> geq(x₁, x₂)
div(x₁, x₂) -> div(x₁, x₂)
if(x₁, x₂, x₃) -> if(x₁, x₂, x₃)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pair:

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

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

         ↳AFS

Dependency Pair:

GEQ(s(X), s(Y)) -> GEQ(X, Y)

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

The following dependency pair can be strictly oriented:

GEQ(s(X), s(Y)) -> GEQ(X, Y)

The following rules can be oriented:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

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

div > geq > {true, false}
div > 0 > {true, false}
div > if > {true, false}
div > s > {true, false}
GEQ > {true, false}
minus > {true, false}

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

GEQ(x₁, x₂) -> GEQ(x₁, x₂)
s(x₁) -> s(x₁)
minus(x₁, x₂) -> x₁
geq(x₁, x₂) -> geq(x₁, x₂)
div(x₁, x₂) -> div(x₁, x₂)
if(x₁, x₂, x₃) -> if(x₁, x₂, x₃)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

Dependency Pair:

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

Dependency Pair:

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

The following dependency pair can be strictly oriented:

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))

The following rules can be oriented:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

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

DIV > 0
{geq, div, false} > true > 0
{geq, div, false} > if > 0
{geq, div, false} > s > 0
minus > 0

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

DIV(x₁, x₂) -> DIV(x₁, x₂)
s(x₁) -> s(x₁)
minus(x₁, x₂) -> x₁
geq(x₁, x₂) -> geq(x₁, x₂)
div(x₁, x₂) -> div(x₁, x₂)
if(x₁, x₂, x₃) -> if(x₁, x₂, x₃)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

Dependency Pair:

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:08 minutes