Termination Proof using AProVE

Term Rewriting System R:
[Y, X]
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

LE(s(X), s(Y)) -> LE(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)
MINUS(s(X), Y) -> LE(s(X), Y)
IFMINUS(false, s(X), Y) -> MINUS(X, Y)
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Remaining

Dependency Pair:

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

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LE(s(X), s(Y)) -> LE(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:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Remaining

Dependency Pair:

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Narrowing Transformation

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)
MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Instantiation Transformation

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)
IFMINUS(false, s(X), Y) -> MINUS(X, Y)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

IFMINUS(false, s(X'), 0) -> MINUS(X', 0)
IFMINUS(false, s(X'), s(Y'''')) -> MINUS(X', s(Y''''))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Inst

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

IFMINUS(false, s(X'), s(Y'''')) -> MINUS(X', s(Y''''))
MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

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

IFMINUS(false, s(X'), s(Y'''')) -> MINUS(X', s(Y''''))

one new Dependency Pair is created:

IFMINUS(false, s(s(X'''')), s(Y''''')) -> MINUS(s(X''''), s(Y'''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Inst

...

               →DP Problem 8

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

IFMINUS(false, s(s(X'''')), s(Y''''')) -> MINUS(s(X''''), s(Y'''''))
MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

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

MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))

one new Dependency Pair is created:

MINUS(s(s(X'''''')), s(Y''')) -> IFMINUS(le(s(X''''''), Y'''), s(s(X'''''')), s(Y'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
MINUS(s(s(X'''''')), s(Y''')) -> IFMINUS(le(s(X''''''), Y'''), s(s(X'''''')), s(Y'''))
IFMINUS(false, s(s(X'''')), s(Y''''')) -> MINUS(s(X''''), s(Y'''''))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Inst

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

IFMINUS(false, s(X'), 0) -> MINUS(X', 0)
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

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

IFMINUS(false, s(X'), 0) -> MINUS(X', 0)

one new Dependency Pair is created:

IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Inst

...

               →DP Problem 9

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

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

MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)

one new Dependency Pair is created:

MINUS(s(s(X'''''')), 0) -> IFMINUS(false, s(s(X'''''')), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Inst

...

               →DP Problem 11

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

MINUS(s(s(X'''''')), 0) -> IFMINUS(false, s(s(X'''''')), 0)
IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)

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:

IFMINUS(x₁, x₂, x₃) -> x₂
s(x₁) -> s(x₁)
MINUS(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Inst

...

               →DP Problem 12

                 ↳Dependency Graph

       →DP Problem 3

         ↳Remaining

Dependency Pair:

MINUS(s(s(X'''''')), 0) -> IFMINUS(false, s(s(X'''''')), 0)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
MINUS(s(s(X'''''')), s(Y''')) -> IFMINUS(le(s(X''''''), Y'''), s(s(X'''''')), s(Y'''))
IFMINUS(false, s(s(X'''')), s(Y''''')) -> MINUS(s(X''''), s(Y'''''))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:
innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes