YES
The TRS could be proven terminating. The proof took 1477 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (84ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (868ms).
| | – Problem 3 was processed with processor DependencyGraph (3ms).
| | | – Problem 4 was processed with processor PolynomialLinearRange4iUR (322ms).
| | | | – Problem 5 was processed with processor DependencyGraph (1ms).
| | | | | – Problem 6 was processed with processor PolynomialLinearRange4iUR (15ms).
| | | | | | – Problem 7 was processed with processor PolynomialLinearRange4iUR (35ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| a__hd#(cons(X, Y)) | → | mark#(X) | | mark#(incr(X)) | → | a__incr#(mark(X)) |
| a__nats# | → | a__adx#(a__zeros) | | mark#(hd(X)) | → | mark#(X) |
| mark#(zeros) | → | a__zeros# | | mark#(nats) | → | a__nats# |
| mark#(incr(X)) | → | mark#(X) | | mark#(tl(X)) | → | a__tl#(mark(X)) |
| mark#(adx(X)) | → | a__adx#(mark(X)) | | mark#(hd(X)) | → | a__hd#(mark(X)) |
| mark#(tl(X)) | → | mark#(X) | | mark#(adx(X)) | → | mark#(X) |
| a__adx#(cons(X, Y)) | → | a__incr#(cons(X, adx(Y))) | | a__tl#(cons(X, Y)) | → | mark#(Y) |
| a__nats# | → | a__zeros# |
Rewrite Rules
| a__nats | → | a__adx(a__zeros) | | a__zeros | → | cons(0, zeros) |
| a__incr(cons(X, Y)) | → | cons(s(X), incr(Y)) | | a__adx(cons(X, Y)) | → | a__incr(cons(X, adx(Y))) |
| a__hd(cons(X, Y)) | → | mark(X) | | a__tl(cons(X, Y)) | → | mark(Y) |
| mark(nats) | → | a__nats | | mark(adx(X)) | → | a__adx(mark(X)) |
| mark(zeros) | → | a__zeros | | mark(incr(X)) | → | a__incr(mark(X)) |
| mark(hd(X)) | → | a__hd(mark(X)) | | mark(tl(X)) | → | a__tl(mark(X)) |
| mark(cons(X1, X2)) | → | cons(X1, X2) | | mark(0) | → | 0 |
| mark(s(X)) | → | s(X) | | a__nats | → | nats |
| a__adx(X) | → | adx(X) | | a__zeros | → | zeros |
| a__incr(X) | → | incr(X) | | a__hd(X) | → | hd(X) |
| a__tl(X) | → | tl(X) |
Original Signature
Termination of terms over the following signature is verified: a__zeros, tl, hd, mark, a__nats, a__hd, nats, 0, s, a__incr, zeros, a__tl, adx, a__adx, incr, cons
Strategy
The following SCCs where found
| mark#(tl(X)) → mark#(X) | mark#(adx(X)) → mark#(X) |
| a__hd#(cons(X, Y)) → mark#(X) | mark#(hd(X)) → mark#(X) |
| a__tl#(cons(X, Y)) → mark#(Y) | mark#(incr(X)) → mark#(X) |
| mark#(tl(X)) → a__tl#(mark(X)) | mark#(hd(X)) → a__hd#(mark(X)) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| mark#(tl(X)) | → | mark#(X) | | mark#(adx(X)) | → | mark#(X) |
| a__hd#(cons(X, Y)) | → | mark#(X) | | mark#(hd(X)) | → | mark#(X) |
| a__tl#(cons(X, Y)) | → | mark#(Y) | | mark#(incr(X)) | → | mark#(X) |
| mark#(tl(X)) | → | a__tl#(mark(X)) | | mark#(hd(X)) | → | a__hd#(mark(X)) |
Rewrite Rules
| a__nats | → | a__adx(a__zeros) | | a__zeros | → | cons(0, zeros) |
| a__incr(cons(X, Y)) | → | cons(s(X), incr(Y)) | | a__adx(cons(X, Y)) | → | a__incr(cons(X, adx(Y))) |
| a__hd(cons(X, Y)) | → | mark(X) | | a__tl(cons(X, Y)) | → | mark(Y) |
| mark(nats) | → | a__nats | | mark(adx(X)) | → | a__adx(mark(X)) |
| mark(zeros) | → | a__zeros | | mark(incr(X)) | → | a__incr(mark(X)) |
| mark(hd(X)) | → | a__hd(mark(X)) | | mark(tl(X)) | → | a__tl(mark(X)) |
| mark(cons(X1, X2)) | → | cons(X1, X2) | | mark(0) | → | 0 |
| mark(s(X)) | → | s(X) | | a__nats | → | nats |
| a__adx(X) | → | adx(X) | | a__zeros | → | zeros |
| a__incr(X) | → | incr(X) | | a__hd(X) | → | hd(X) |
| a__tl(X) | → | tl(X) |
Original Signature
Termination of terms over the following signature is verified: a__zeros, tl, hd, mark, a__nats, a__hd, nats, 0, s, a__incr, zeros, a__tl, adx, a__adx, incr, cons
Strategy
Polynomial Interpretation
- 0: 0
- a__adx(x): x
- a__hd(x): x + 1
- a__hd#(x): x + 1
- a__incr(x): x
- a__nats: 0
- a__tl(x): 2x + 1
- a__tl#(x): 2x
- a__zeros: 0
- adx(x): x
- cons(x,y): y + 2x
- hd(x): x + 1
- incr(x): x
- mark(x): 2x + 1
- mark#(x): 2x
- nats: 0
- s(x): 0
- tl(x): 2x + 1
- zeros: 0
Improved Usable rules
| mark(s(X)) | → | s(X) | | mark(0) | → | 0 |
| a__tl(X) | → | tl(X) | | a__adx(cons(X, Y)) | → | a__incr(cons(X, adx(Y))) |
| mark(cons(X1, X2)) | → | cons(X1, X2) | | a__incr(cons(X, Y)) | → | cons(s(X), incr(Y)) |
| a__adx(X) | → | adx(X) | | a__zeros | → | zeros |
| a__hd(X) | → | hd(X) | | mark(hd(X)) | → | a__hd(mark(X)) |
| a__nats | → | a__adx(a__zeros) | | mark(nats) | → | a__nats |
| mark(zeros) | → | a__zeros | | mark(tl(X)) | → | a__tl(mark(X)) |
| mark(incr(X)) | → | a__incr(mark(X)) | | a__nats | → | nats |
| a__tl(cons(X, Y)) | → | mark(Y) | | a__incr(X) | → | incr(X) |
| a__zeros | → | cons(0, zeros) | | mark(adx(X)) | → | a__adx(mark(X)) |
| a__hd(cons(X, Y)) | → | mark(X) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| mark#(tl(X)) | → | mark#(X) | | a__hd#(cons(X, Y)) | → | mark#(X) |
| mark#(hd(X)) | → | mark#(X) |
Problem 3: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| mark#(adx(X)) | → | mark#(X) | | a__tl#(cons(X, Y)) | → | mark#(Y) |
| mark#(incr(X)) | → | mark#(X) | | mark#(tl(X)) | → | a__tl#(mark(X)) |
| mark#(hd(X)) | → | a__hd#(mark(X)) |
Rewrite Rules
| a__nats | → | a__adx(a__zeros) | | a__zeros | → | cons(0, zeros) |
| a__incr(cons(X, Y)) | → | cons(s(X), incr(Y)) | | a__adx(cons(X, Y)) | → | a__incr(cons(X, adx(Y))) |
| a__hd(cons(X, Y)) | → | mark(X) | | a__tl(cons(X, Y)) | → | mark(Y) |
| mark(nats) | → | a__nats | | mark(adx(X)) | → | a__adx(mark(X)) |
| mark(zeros) | → | a__zeros | | mark(incr(X)) | → | a__incr(mark(X)) |
| mark(hd(X)) | → | a__hd(mark(X)) | | mark(tl(X)) | → | a__tl(mark(X)) |
| mark(cons(X1, X2)) | → | cons(X1, X2) | | mark(0) | → | 0 |
| mark(s(X)) | → | s(X) | | a__nats | → | nats |
| a__adx(X) | → | adx(X) | | a__zeros | → | zeros |
| a__incr(X) | → | incr(X) | | a__hd(X) | → | hd(X) |
| a__tl(X) | → | tl(X) |
Original Signature
Termination of terms over the following signature is verified: a__zeros, tl, hd, mark, a__nats, a__hd, nats, 0, s, a__incr, zeros, a__tl, a__adx, adx, incr, cons
Strategy
The following SCCs where found
| mark#(adx(X)) → mark#(X) | a__tl#(cons(X, Y)) → mark#(Y) |
| mark#(incr(X)) → mark#(X) | mark#(tl(X)) → a__tl#(mark(X)) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| mark#(adx(X)) | → | mark#(X) | | a__tl#(cons(X, Y)) | → | mark#(Y) |
| mark#(incr(X)) | → | mark#(X) | | mark#(tl(X)) | → | a__tl#(mark(X)) |
Rewrite Rules
| a__nats | → | a__adx(a__zeros) | | a__zeros | → | cons(0, zeros) |
| a__incr(cons(X, Y)) | → | cons(s(X), incr(Y)) | | a__adx(cons(X, Y)) | → | a__incr(cons(X, adx(Y))) |
| a__hd(cons(X, Y)) | → | mark(X) | | a__tl(cons(X, Y)) | → | mark(Y) |
| mark(nats) | → | a__nats | | mark(adx(X)) | → | a__adx(mark(X)) |
| mark(zeros) | → | a__zeros | | mark(incr(X)) | → | a__incr(mark(X)) |
| mark(hd(X)) | → | a__hd(mark(X)) | | mark(tl(X)) | → | a__tl(mark(X)) |
| mark(cons(X1, X2)) | → | cons(X1, X2) | | mark(0) | → | 0 |
| mark(s(X)) | → | s(X) | | a__nats | → | nats |
| a__adx(X) | → | adx(X) | | a__zeros | → | zeros |
| a__incr(X) | → | incr(X) | | a__hd(X) | → | hd(X) |
| a__tl(X) | → | tl(X) |
Original Signature
Termination of terms over the following signature is verified: a__zeros, tl, hd, mark, a__nats, a__hd, nats, 0, s, a__incr, zeros, a__tl, a__adx, adx, incr, cons
Strategy
Polynomial Interpretation
- 0: 0
- a__adx(x): x
- a__hd(x): 2x
- a__incr(x): x
- a__nats: 0
- a__tl(x): 3x + 1
- a__tl#(x): 3x
- a__zeros: 0
- adx(x): x
- cons(x,y): 3y + x
- hd(x): 2x
- incr(x): x
- mark(x): 2x
- mark#(x): 3x
- nats: 0
- s(x): 0
- tl(x): 3x + 1
- zeros: 0
Improved Usable rules
| mark(s(X)) | → | s(X) | | mark(0) | → | 0 |
| a__tl(X) | → | tl(X) | | a__adx(cons(X, Y)) | → | a__incr(cons(X, adx(Y))) |
| mark(cons(X1, X2)) | → | cons(X1, X2) | | a__incr(cons(X, Y)) | → | cons(s(X), incr(Y)) |
| a__adx(X) | → | adx(X) | | a__zeros | → | zeros |
| a__hd(X) | → | hd(X) | | mark(hd(X)) | → | a__hd(mark(X)) |
| a__nats | → | a__adx(a__zeros) | | mark(nats) | → | a__nats |
| mark(zeros) | → | a__zeros | | mark(tl(X)) | → | a__tl(mark(X)) |
| mark(incr(X)) | → | a__incr(mark(X)) | | a__nats | → | nats |
| a__tl(cons(X, Y)) | → | mark(Y) | | a__incr(X) | → | incr(X) |
| a__zeros | → | cons(0, zeros) | | mark(adx(X)) | → | a__adx(mark(X)) |
| a__hd(cons(X, Y)) | → | mark(X) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| mark#(tl(X)) | → | a__tl#(mark(X)) |
Problem 5: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| mark#(adx(X)) | → | mark#(X) | | a__tl#(cons(X, Y)) | → | mark#(Y) |
| mark#(incr(X)) | → | mark#(X) |
Rewrite Rules
| a__nats | → | a__adx(a__zeros) | | a__zeros | → | cons(0, zeros) |
| a__incr(cons(X, Y)) | → | cons(s(X), incr(Y)) | | a__adx(cons(X, Y)) | → | a__incr(cons(X, adx(Y))) |
| a__hd(cons(X, Y)) | → | mark(X) | | a__tl(cons(X, Y)) | → | mark(Y) |
| mark(nats) | → | a__nats | | mark(adx(X)) | → | a__adx(mark(X)) |
| mark(zeros) | → | a__zeros | | mark(incr(X)) | → | a__incr(mark(X)) |
| mark(hd(X)) | → | a__hd(mark(X)) | | mark(tl(X)) | → | a__tl(mark(X)) |
| mark(cons(X1, X2)) | → | cons(X1, X2) | | mark(0) | → | 0 |
| mark(s(X)) | → | s(X) | | a__nats | → | nats |
| a__adx(X) | → | adx(X) | | a__zeros | → | zeros |
| a__incr(X) | → | incr(X) | | a__hd(X) | → | hd(X) |
| a__tl(X) | → | tl(X) |
Original Signature
Termination of terms over the following signature is verified: a__zeros, tl, hd, mark, a__nats, a__hd, nats, 0, s, a__incr, zeros, a__tl, adx, a__adx, incr, cons
Strategy
The following SCCs where found
| mark#(adx(X)) → mark#(X) | mark#(incr(X)) → mark#(X) |
Problem 6: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| mark#(adx(X)) | → | mark#(X) | | mark#(incr(X)) | → | mark#(X) |
Rewrite Rules
| a__nats | → | a__adx(a__zeros) | | a__zeros | → | cons(0, zeros) |
| a__incr(cons(X, Y)) | → | cons(s(X), incr(Y)) | | a__adx(cons(X, Y)) | → | a__incr(cons(X, adx(Y))) |
| a__hd(cons(X, Y)) | → | mark(X) | | a__tl(cons(X, Y)) | → | mark(Y) |
| mark(nats) | → | a__nats | | mark(adx(X)) | → | a__adx(mark(X)) |
| mark(zeros) | → | a__zeros | | mark(incr(X)) | → | a__incr(mark(X)) |
| mark(hd(X)) | → | a__hd(mark(X)) | | mark(tl(X)) | → | a__tl(mark(X)) |
| mark(cons(X1, X2)) | → | cons(X1, X2) | | mark(0) | → | 0 |
| mark(s(X)) | → | s(X) | | a__nats | → | nats |
| a__adx(X) | → | adx(X) | | a__zeros | → | zeros |
| a__incr(X) | → | incr(X) | | a__hd(X) | → | hd(X) |
| a__tl(X) | → | tl(X) |
Original Signature
Termination of terms over the following signature is verified: a__zeros, tl, hd, mark, a__nats, a__hd, nats, 0, s, a__incr, zeros, a__tl, adx, a__adx, incr, cons
Strategy
Polynomial Interpretation
- 0: 0
- a__adx(x): 0
- a__hd(x): 0
- a__incr(x): 0
- a__nats: 0
- a__tl(x): 0
- a__zeros: 0
- adx(x): 3x
- cons(x,y): 0
- hd(x): 0
- incr(x): x + 1
- mark(x): 0
- mark#(x): 3x
- nats: 0
- s(x): 0
- tl(x): 0
- zeros: 0
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| mark#(incr(X)) | → | mark#(X) |
Problem 7: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| a__nats | → | a__adx(a__zeros) | | a__zeros | → | cons(0, zeros) |
| a__incr(cons(X, Y)) | → | cons(s(X), incr(Y)) | | a__adx(cons(X, Y)) | → | a__incr(cons(X, adx(Y))) |
| a__hd(cons(X, Y)) | → | mark(X) | | a__tl(cons(X, Y)) | → | mark(Y) |
| mark(nats) | → | a__nats | | mark(adx(X)) | → | a__adx(mark(X)) |
| mark(zeros) | → | a__zeros | | mark(incr(X)) | → | a__incr(mark(X)) |
| mark(hd(X)) | → | a__hd(mark(X)) | | mark(tl(X)) | → | a__tl(mark(X)) |
| mark(cons(X1, X2)) | → | cons(X1, X2) | | mark(0) | → | 0 |
| mark(s(X)) | → | s(X) | | a__nats | → | nats |
| a__adx(X) | → | adx(X) | | a__zeros | → | zeros |
| a__incr(X) | → | incr(X) | | a__hd(X) | → | hd(X) |
| a__tl(X) | → | tl(X) |
Original Signature
Termination of terms over the following signature is verified: a__zeros, tl, hd, mark, a__nats, a__hd, nats, 0, s, a__incr, zeros, a__tl, a__adx, adx, incr, cons
Strategy
Polynomial Interpretation
- 0: 0
- a__adx(x): 0
- a__hd(x): 0
- a__incr(x): 0
- a__nats: 0
- a__tl(x): 0
- a__zeros: 0
- adx(x): x + 2
- cons(x,y): 0
- hd(x): 0
- incr(x): 0
- mark(x): 0
- mark#(x): x + 1
- nats: 0
- s(x): 0
- tl(x): 0
- zeros: 0
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed: