Termination Proof using AProVE

Term Rewriting System R:
[X, Y, N, M]
eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

EQ(s(X), s(Y)) -> EQ(X, Y)
RM(N, add(M, X)) -> IFRM(eq(N, M), N, add(M, X))
RM(N, add(M, X)) -> EQ(N, M)
IFRM(true, N, add(M, X)) -> RM(N, X)
IFRM(false, N, add(M, X)) -> RM(N, X)
PURGE(add(N, X)) -> PURGE(rm(N, X))
PURGE(add(N, X)) -> RM(N, X)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Remaining

Dependency Pair:

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

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

EQ(x₁, x₂) -> EQ(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:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

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:

IFRM(false, N, add(M, X)) -> RM(N, X)
IFRM(true, N, add(M, X)) -> RM(N, X)
RM(N, add(M, X)) -> IFRM(eq(N, M), N, add(M, X))

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Strategy:

innermost

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

RM(N, add(M, X)) -> IFRM(eq(N, M), N, add(M, X))

four new Dependency Pairs are created:

RM(0, add(0, X)) -> IFRM(true, 0, add(0, X))
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))
RM(s(X''), add(0, X)) -> IFRM(false, s(X''), add(0, X))
RM(s(X''), add(s(Y'), X)) -> IFRM(eq(X'', Y'), s(X''), add(s(Y'), X))

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:

RM(s(X''), add(s(Y'), X)) -> IFRM(eq(X'', Y'), s(X''), add(s(Y'), X))
RM(s(X''), add(0, X)) -> IFRM(false, s(X''), add(0, X))
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))
IFRM(true, N, add(M, X)) -> RM(N, X)
RM(0, add(0, X)) -> IFRM(true, 0, add(0, X))
IFRM(false, N, add(M, X)) -> RM(N, X)

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Strategy:

innermost

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

IFRM(true, N, add(M, X)) -> RM(N, X)

two new Dependency Pairs are created:

IFRM(true, 0, add(0, X'')) -> RM(0, X'')
IFRM(true, s(X''''), add(s(Y'''), X'')) -> RM(s(X''''), X'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Inst

...

               →DP Problem 6

                 ↳Instantiation Transformation

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

IFRM(true, s(X''''), add(s(Y'''), X'')) -> RM(s(X''''), X'')
RM(s(X''), add(0, X)) -> IFRM(false, s(X''), add(0, X))
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))
IFRM(true, 0, add(0, X'')) -> RM(0, X'')
RM(0, add(0, X)) -> IFRM(true, 0, add(0, X))
IFRM(false, N, add(M, X)) -> RM(N, X)
RM(s(X''), add(s(Y'), X)) -> IFRM(eq(X'', Y'), s(X''), add(s(Y'), X))

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Strategy:

innermost

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

IFRM(false, N, add(M, X)) -> RM(N, X)

three new Dependency Pairs are created:

IFRM(false, 0, add(s(X''''), X'')) -> RM(0, X'')
IFRM(false, s(X''''), add(0, X'')) -> RM(s(X''''), X'')
IFRM(false, s(X''''), add(s(Y'''), X'')) -> RM(s(X''''), X'')

The transformation is resulting in two 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:
IFRM(false, s(X''''), add(s(Y'''), X'')) -> RM(s(X''''), X'')
RM(s(X''), add(s(Y'), X)) -> IFRM(eq(X'', Y'), s(X''), add(s(Y'), X))
IFRM(false, s(X''''), add(0, X'')) -> RM(s(X''''), X'')
RM(s(X''), add(0, X)) -> IFRM(false, s(X''), add(0, X))
IFRM(true, s(X''''), add(s(Y'''), X'')) -> RM(s(X''''), X'')

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Strategy:
innermost
Dependency Pair:
PURGE(add(N, X)) -> PURGE(rm(N, X))

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Inst

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

IFRM(true, 0, add(0, X'')) -> RM(0, X'')
RM(0, add(0, X)) -> IFRM(true, 0, add(0, X))
IFRM(false, 0, add(s(X''''), X'')) -> RM(0, X'')
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IFRM(true, 0, add(0, X'')) -> RM(0, X'')
IFRM(false, 0, add(s(X''''), X'')) -> RM(0, X'')

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:

RM(x₁, x₂) -> x₂
add(x₁, x₂) -> add(x₁, x₂)
IFRM(x₁, x₂, x₃) -> x₃
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Inst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

RM(0, add(0, X)) -> IFRM(true, 0, add(0, X))
RM(0, add(s(X''), X)) -> IFRM(false, 0, add(s(X''), X))

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

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:
IFRM(false, s(X''''), add(s(Y'''), X'')) -> RM(s(X''''), X'')
RM(s(X''), add(s(Y'), X)) -> IFRM(eq(X'', Y'), s(X''), add(s(Y'), X))
IFRM(false, s(X''''), add(0, X'')) -> RM(s(X''''), X'')
RM(s(X''), add(0, X)) -> IFRM(false, s(X''), add(0, X))
IFRM(true, s(X''''), add(s(Y'''), X'')) -> RM(s(X''''), X'')

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Strategy:
innermost
Dependency Pair:
PURGE(add(N, X)) -> PURGE(rm(N, X))

Rules:

eq(0, 0) -> true
eq(0, s(X)) -> false
eq(s(X), 0) -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Strategy:
innermost

Innermost Termination of R could not be shown.
Duration:
0:03 minutes