Termination Proof using AProVE

Term Rewriting System R:
[t, n, x, a, b, c]
g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FOLDB(t, s(n)) -> F(foldB(t, n), B)
FOLDB(t, s(n)) -> FOLDB(t, n)
FOLDC(t, s(n)) -> F(foldC(t, n), C)
FOLDC(t, s(n)) -> FOLDC(t, n)
F(t, x) -> F'(t, g(x))
F(t, x) -> G(x)
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F'(triple(a, b, c), A) -> F''(foldB(triple(s(a), 0, c), b))
F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
FOLD(t, x, s(n)) -> F(fold(t, x, n), x)
FOLD(t, x, s(n)) -> FOLD(t, x, n)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

Dependency Pairs:

FOLDB(t, s(n)) -> FOLDB(t, n)
F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
FOLDC(t, s(n)) -> FOLDC(t, n)
FOLDC(t, s(n)) -> F(foldC(t, n), C)
F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
F'(triple(a, b, c), A) -> F''(foldB(triple(s(a), 0, c), b))
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, x) -> F'(t, g(x))
FOLDB(t, s(n)) -> F(foldB(t, n), B)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳Modular Removal of Rules

       →DP Problem 2

         ↳UsableRules

Dependency Pairs:

Rules:

g(C) -> A
g(B) -> B
g(A) -> A
g(C) -> B
g(B) -> A
g(C) -> C
foldC(t, s(n)) -> f(foldC(t, n), C)
foldC(t, 0) -> t
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
foldB(t, s(n)) -> f(foldB(t, n), B)
foldB(t, 0) -> t
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

Strategy:

innermost

We have the following set of usable rules:

g(C) -> A
g(B) -> B
g(A) -> A
g(C) -> B
g(B) -> A
g(C) -> C
foldC(t, s(n)) -> f(foldC(t, n), C)
foldC(t, 0) -> t
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
foldB(t, s(n)) -> f(foldB(t, n), B)
foldB(t, 0) -> t
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(foldB(x₁, x₂)) = x₁ + x₂

POL(FOLDC(x₁, x₂)) = x₁ + x₂

POL(triple(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(f'(x₁, x₂)) = 1 + x₁ + x₂

POL(f(x₁, x₂)) = 1 + x₁ + x₂

POL(F(x₁, x₂)) = 1 + x₁ + x₂

POL(foldC(x₁, x₂)) = x₁ + x₂

POL(FOLDB(x₁, x₂)) = x₁ + x₂

POL(F''(x₁)) = x₁

POL(0) = 0

POL(C) = 0

POL(g(x₁)) = x₁

POL(B) = 0

POL(F'(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = 1 + x₁

POL(f''(x₁)) = x₁

POL(A) = 0

We have the following set D of usable symbols: {foldB, FOLDC, triple, f', F, f, foldC, FOLDB, F'', C, 0, B, g, F', s, f'', A}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

FOLDB(t, s(n)) -> FOLDB(t, n)
FOLDC(t, s(n)) -> FOLDC(t, n)

No Rules can be deleted.

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳MRR

...

               →DP Problem 4

                 ↳Negative Polynomial Order

       →DP Problem 2

         ↳UsableRules

Dependency Pairs:

F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
FOLDC(t, s(n)) -> F(foldC(t, n), C)
F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
F'(triple(a, b, c), A) -> F''(foldB(triple(s(a), 0, c), b))
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, x) -> F'(t, g(x))
FOLDB(t, s(n)) -> F(foldB(t, n), B)

Rules:

g(C) -> A
g(B) -> B
g(A) -> A
g(C) -> B
g(B) -> A
g(C) -> C
foldC(t, s(n)) -> f(foldC(t, n), C)
foldC(t, 0) -> t
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
foldB(t, s(n)) -> f(foldB(t, n), B)
foldB(t, 0) -> t
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)

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

g(C) -> A
foldC(t, s(n)) -> f(foldC(t, n), C)
g(B) -> B
foldC(t, 0) -> t
g(A) -> A
g(B) -> A
g(C) -> B
f(t, x) -> f'(t, g(x))
foldB(t, s(n)) -> f(foldB(t, n), B)
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
foldB(t, 0) -> t
g(C) -> C
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))

Used ordering:
Polynomial Order with Interpretation:

POL( F'(x₁, x₂) ) = x₁ + 1

POL( triple(x₁, ..., x₃) ) = x₂

POL( FOLDB(x₁, x₂) ) = x₁ + x₂

POL( 0 ) = 0

POL( FOLDC(x₁, x₂) ) = x₁ + 1

POL( F(x₁, x₂) ) = x₁ + 1

POL( foldC(x₁, x₂) ) = x₁

POL( s(x₁) ) = 1

POL( foldB(x₁, x₂) ) = x₁

POL( F''(x₁) ) = x₁ + 1

POL( g(x₁) ) = 0

POL( A ) = 0

POL( f(x₁, x₂) ) = x₁

POL( B ) = 0

POL( f'(x₁, x₂) ) = x₁

POL( f''(x₁) ) = x₁

POL( C ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳MRR

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳UsableRules

Dependency Pairs:

FOLDC(t, s(n)) -> F(foldC(t, n), C)
F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
F'(triple(a, b, c), A) -> F''(foldB(triple(s(a), 0, c), b))
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, x) -> F'(t, g(x))
FOLDB(t, s(n)) -> F(foldB(t, n), B)

Rules:

g(C) -> A
g(B) -> B
g(A) -> A
g(C) -> B
g(B) -> A
g(C) -> C
foldC(t, s(n)) -> f(foldC(t, n), C)
foldC(t, 0) -> t
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
foldB(t, s(n)) -> f(foldB(t, n), B)
foldB(t, 0) -> t
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳MRR

...

               →DP Problem 6

                 ↳Negative Polynomial Order

       →DP Problem 2

         ↳UsableRules

Dependency Pairs:

F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
F'(triple(a, b, c), A) -> F''(foldB(triple(s(a), 0, c), b))
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, x) -> F'(t, g(x))
FOLDC(t, s(n)) -> F(foldC(t, n), C)

Rules:

g(C) -> A
g(B) -> B
g(A) -> A
g(C) -> B
g(B) -> A
g(C) -> C
foldC(t, s(n)) -> f(foldC(t, n), C)
foldC(t, 0) -> t
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
foldB(t, s(n)) -> f(foldB(t, n), B)
foldB(t, 0) -> t
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)

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

g(C) -> A
foldC(t, s(n)) -> f(foldC(t, n), C)
g(B) -> B
foldC(t, 0) -> t
g(A) -> A
g(B) -> A
g(C) -> B
f(t, x) -> f'(t, g(x))
foldB(t, s(n)) -> f(foldB(t, n), B)
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
foldB(t, 0) -> t
g(C) -> C
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))

Used ordering:
Polynomial Order with Interpretation:

POL( F''(x₁) ) = x₁ + 1

POL( triple(x₁, ..., x₃) ) = x₃

POL( FOLDC(x₁, x₂) ) = x₁ + x₂

POL( 0 ) = 0

POL( s(x₁) ) = x₁ + 1

POL( F(x₁, x₂) ) = x₁ + 1

POL( foldC(x₁, x₂) ) = x₁ + x₂

POL( F'(x₁, x₂) ) = x₁ + 1

POL( foldB(x₁, x₂) ) = x₁

POL( g(x₁) ) = x₁

POL( C ) = 1

POL( A ) = 0

POL( f(x₁, x₂) ) = x₁ + x₂

POL( B ) = 0

POL( f'(x₁, x₂) ) = x₁ + x₂

POL( f''(x₁) ) = x₁

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳MRR

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳UsableRules

Dependency Pairs:

F'(triple(a, b, c), A) -> F''(foldB(triple(s(a), 0, c), b))
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, x) -> F'(t, g(x))
FOLDC(t, s(n)) -> F(foldC(t, n), C)

Rules:

g(C) -> A
g(B) -> B
g(A) -> A
g(C) -> B
g(B) -> A
g(C) -> C
foldC(t, s(n)) -> f(foldC(t, n), C)
foldC(t, 0) -> t
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
foldB(t, s(n)) -> f(foldB(t, n), B)
foldB(t, 0) -> t
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳MRR

...

               →DP Problem 8

                 ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

Dependency Pairs:

F(t, x) -> F'(t, g(x))
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)

Rules:

g(C) -> A
g(B) -> B
g(A) -> A
g(C) -> B
g(B) -> A
g(C) -> C
foldC(t, s(n)) -> f(foldC(t, n), C)
foldC(t, 0) -> t
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
foldB(t, s(n)) -> f(foldB(t, n), B)
foldB(t, 0) -> t
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

Strategy:

innermost

As we are in the innermost case, we can delete all 9 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳MRR

...

               →DP Problem 9

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳UsableRules

Dependency Pairs:

F(t, x) -> F'(t, g(x))
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)

Rules:

g(C) -> A
g(B) -> B
g(C) -> C
g(A) -> A
g(C) -> B
g(B) -> A

Strategy:

innermost

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

F(t, x) -> F'(t, g(x))

six new Dependency Pairs are created:

F(t, C) -> F'(t, A)
F(t, B) -> F'(t, B)
F(t, C) -> F'(t, C)
F(t, A) -> F'(t, A)
F(t, C) -> F'(t, B)
F(t, B) -> F'(t, A)

The transformation is resulting in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

Dependency Pair:

FOLD(t, x, s(n)) -> FOLD(t, x, n)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Strategy:

innermost

As we are in the innermost case, we can delete all 17 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 10

             ↳Size-Change Principle

Dependency Pair:

FOLD(t, x, s(n)) -> FOLD(t, x, n)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

FOLD(t, x, s(n)) -> FOLD(t, x, n)

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

{1}	,	{1}
1	=	1
2	=	2
3	>	3

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

{1}	,	{1}
1	=	1
2	=	2
3	>	3

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.

Innermost Termination of R successfully shown.
Duration:
0:07 minutes

POL(foldB(x₁, x₂))	= x₁ + x₂
POL(FOLDC(x₁, x₂))	= x₁ + x₂
POL(triple(x₁, x₂, x₃))	= x₁ + x₂ + x₃
POL(f'(x₁, x₂))	= 1 + x₁ + x₂
POL(f(x₁, x₂))	= 1 + x₁ + x₂
POL(F(x₁, x₂))	= 1 + x₁ + x₂
POL(foldC(x₁, x₂))	= x₁ + x₂
POL(FOLDB(x₁, x₂))	= x₁ + x₂
POL(F''(x₁))	= x₁
POL(0)	= 0
POL(C)	= 0
POL(g(x₁))	= x₁
POL(B)	= 0
POL(F'(x₁, x₂))	= 1 + x₁ + x₂
POL(s(x₁))	= 1 + x₁
POL(f''(x₁))	= x₁
POL(A)	= 0