YES

The TRS could be proven terminating. The proof took 508 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (44ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4 (101ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4 (171ms).
 |    | – Problem 4 was processed with processor DependencyGraph (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

*top*_0#(f_1(i_1(x)))T(x)i_1#(x)h_1#(x)
h_1#(x)f_1#(h_1(x))T(i_1(x))i_1#(x)
T(i_1(x_1))T(x_1)f_0#(f_1(i_1(x)))f_1#(x)
*top*_0#(f_1(i_1(x)))*top*_0#(x)T(h_1(x_1))T(x_1)
f_1#(h_1(x))f_1#(i_1(x))T(h_1(x))h_1#(x)
f_1#(i_1(x))T(x)

Rewrite Rules

f_1(h_1(x))f_1(i_1(x))h_1(x)f_1(h_1(x))
*top*_0(f_1(i_1(x)))*top*_0(x)f_0(f_1(i_1(x)))f_1(x)
f_1(i_1(x))xi_1(x)h_1(x)

Original Signature

Termination of terms over the following signature is verified: f_0, i_1, h_1, *top*_0, f_1

Strategy

Context-sensitive strategy:
μ(h_1#) = μ(T) = μ(i_1) = μ(f_1#) = μ(h_1) = μ(f_1) = μ(i_1#) = ∅
μ(f_0) = μ(*top*_0) = μ(f_0#) = μ(*top*_0#) = {1}


The following SCCs where found

i_1#(x) → h_1#(x)h_1#(x) → f_1#(h_1(x))
T(i_1(x)) → i_1#(x)T(i_1(x_1)) → T(x_1)
T(h_1(x_1)) → T(x_1)f_1#(h_1(x)) → f_1#(i_1(x))
T(h_1(x)) → h_1#(x)f_1#(i_1(x)) → T(x)

*top*_0#(f_1(i_1(x))) → *top*_0#(x)

Problem 2: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

*top*_0#(f_1(i_1(x)))*top*_0#(x)

Rewrite Rules

f_1(h_1(x))f_1(i_1(x))h_1(x)f_1(h_1(x))
*top*_0(f_1(i_1(x)))*top*_0(x)f_0(f_1(i_1(x)))f_1(x)
f_1(i_1(x))xi_1(x)h_1(x)

Original Signature

Termination of terms over the following signature is verified: f_0, i_1, h_1, *top*_0, f_1

Strategy

Context-sensitive strategy:
μ(T) = μ(h_1#) = μ(i_1) = μ(f_1#) = μ(h_1) = μ(f_1) = μ(i_1#) = ∅
μ(f_0) = μ(*top*_0) = μ(*top*_0#) = μ(f_0#) = {1}


Polynomial Interpretation

Standard Usable rules

f_1(i_1(x))xh_1(x)f_1(h_1(x))
*top*_0(f_1(i_1(x)))*top*_0(x)f_1(h_1(x))f_1(i_1(x))
i_1(x)h_1(x)

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

*top*_0#(f_1(i_1(x)))*top*_0#(x)

Problem 3: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

i_1#(x)h_1#(x)h_1#(x)f_1#(h_1(x))
T(i_1(x))i_1#(x)T(i_1(x_1))T(x_1)
T(h_1(x_1))T(x_1)f_1#(h_1(x))f_1#(i_1(x))
T(h_1(x))h_1#(x)f_1#(i_1(x))T(x)

Rewrite Rules

f_1(h_1(x))f_1(i_1(x))h_1(x)f_1(h_1(x))
*top*_0(f_1(i_1(x)))*top*_0(x)f_0(f_1(i_1(x)))f_1(x)
f_1(i_1(x))xi_1(x)h_1(x)

Original Signature

Termination of terms over the following signature is verified: f_0, i_1, h_1, *top*_0, f_1

Strategy

Context-sensitive strategy:
μ(T) = μ(h_1#) = μ(i_1) = μ(f_1#) = μ(h_1) = μ(f_1) = μ(i_1#) = ∅
μ(f_0) = μ(*top*_0) = μ(*top*_0#) = μ(f_0#) = {1}


Polynomial Interpretation

Standard Usable rules

f_1(i_1(x))xh_1(x)f_1(h_1(x))
*top*_0(f_1(i_1(x)))*top*_0(x)f_1(h_1(x))f_1(i_1(x))
i_1(x)h_1(x)

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

i_1#(x)h_1#(x)T(i_1(x_1))T(x_1)
T(h_1(x_1))T(x_1)T(h_1(x))h_1#(x)
f_1#(i_1(x))T(x)

Problem 4: DependencyGraph



Dependency Pair Problem

Dependency Pairs

T(i_1(x))i_1#(x)h_1#(x)f_1#(h_1(x))
f_1#(h_1(x))f_1#(i_1(x))

Rewrite Rules

f_1(h_1(x))f_1(i_1(x))h_1(x)f_1(h_1(x))
*top*_0(f_1(i_1(x)))*top*_0(x)f_0(f_1(i_1(x)))f_1(x)
f_1(i_1(x))xi_1(x)h_1(x)

Original Signature

Termination of terms over the following signature is verified: f_0, i_1, h_1, *top*_0, f_1

Strategy

Context-sensitive strategy:
μ(h_1#) = μ(T) = μ(i_1) = μ(f_1#) = μ(h_1) = μ(f_1) = μ(i_1#) = ∅
μ(f_0) = μ(*top*_0) = μ(f_0#) = μ(*top*_0#) = {1}


There are no SCCs!