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

         ↳AFS

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

trivial

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

         ↳AFS

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

         ↳AFS

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

trivial

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

         ↳AFS

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

         ↳Argument Filtering and Ordering

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

The following dependency pairs can be strictly oriented:

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)

The following usable rules for innermost can be oriented:

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

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

mark > {true, 0}
{geq, a_{_div}, a_{_geq}, div, A_{_DIV}, false} > {a_{_if}, if} > A_{_IF} > MARK > {true, 0}
{geq, a_{_div}, a_{_geq}, div, A_{_DIV}, false} > s > {true, 0}
minus > {true, 0}
a_{_minus} > {true, 0}

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

A_{_DIV}(x₁, x₂) -> A_{_DIV}(x₁, x₂)
A_{_IF}(x₁, x₂, x₃) -> A_{_IF}(x₁, x₂, x₃)
s(x₁) -> s(x₁)
a_{_geq}(x₁, x₂) -> a_{_geq}(x₁, x₂)
div(x₁, x₂) -> div(x₁, x₂)
minus(x₁, x₂) -> x₁
MARK(x₁) -> MARK(x₁)
if(x₁, x₂, x₃) -> if(x₁, x₂, x₃)
mark(x₁) -> x₁
geq(x₁, x₂) -> geq(x₁, x₂)
a_{_minus}(x₁, x₂) -> x₁
a_{_div}(x₁, x₂) -> a_{_div}(x₁, x₂)
a_{_if}(x₁, x₂, x₃) -> a_{_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:

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.

Innermost Termination of R successfully shown.
Duration:
0:13 minutes