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

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Neg POLO

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

As we are in the innermost case, we can delete all 21 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Neg POLO

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1
2	>	2

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1
2	>	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

s(x₁) -> s(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳Neg POLO

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

As we are in the innermost case, we can delete all 21 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳Size-Change Principle

       →DP Problem 3

         ↳Neg POLO

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1
2	>	2

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1
2	>	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

s(x₁) -> s(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Negative Polynomial Order

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 Pair can be strictly oriented using the given order.

MARK(s(X)) -> MARK(X)

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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

Used ordering:
Polynomial Order with Interpretation:

POL( MARK(x₁) ) = x₁

POL( s(x₁) ) = x₁ + 1

POL( A_{_DIV}(x₁, x₂) ) = x₁

POL( A_{_IF}(x₁, ..., x₃) ) = x₂ + x₃

POL( div(x₁, x₂) ) = x₁

POL( minus(x₁, x₂) ) = 0

POL( 0 ) = 0

POL( if(x₁, ..., x₃) ) = x₁ + x₂ + x₃

POL( mark(x₁) ) = x₁

POL( a_{_if}(x₁, ..., x₃) ) = x₁ + x₂ + x₃

POL( a_{_div}(x₁, x₂) ) = x₁

POL( a_{_geq}(x₁, x₂) ) = 0

POL( a_{_minus}(x₁, x₂) ) = 0

POL( true ) = 0

POL( false ) = 0

POL( geq(x₁, x₂) ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Neg POLO

           →DP Problem 6

             ↳Negative Polynomial Order

Dependency Pairs:

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 using the given order.

MARK(if(X1, X2, X3)) -> MARK(X1)
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)

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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

Used ordering:
Polynomial Order with Interpretation:

POL( MARK(x₁) ) = x₁

POL( if(x₁, ..., x₃) ) = x₁ + x₂ + x₃ + 1

POL( A_{_DIV}(x₁, x₂) ) = 0

POL( A_{_IF}(x₁, ..., x₃) ) = x₂ + x₃

POL( s(x₁) ) = 0

POL( 0 ) = 0

POL( div(x₁, x₂) ) = x₁ + 1

POL( a_{_if}(x₁, ..., x₃) ) = x₁ + x₂ + x₃ + 1

POL( false ) = 0

POL( mark(x₁) ) = x₁

POL( a_{_div}(x₁, x₂) ) = x₁ + 1

POL( a_{_geq}(x₁, x₂) ) = 0

POL( minus(x₁, x₂) ) = 0

POL( a_{_minus}(x₁, x₂) ) = 0

POL( true ) = 0

POL( geq(x₁, x₂) ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Neg POLO

           →DP Problem 6

             ↳Neg POLO

...

               →DP Problem 7

                 ↳Dependency Graph

Dependency Pairs:

A_{_IF}(false, X, Y) -> MARK(Y)
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

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:03 minutes