YES
The TRS could be proven terminating. The proof took 44 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (22ms).
| Problem 2 was processed with processor SubtermCriterion (2ms).
| Problem 3 was processed with processor SubtermCriterion (3ms).
| Problem 4 was processed with processor SubtermCriterion (1ms).
| Problem 5 was processed with processor SubtermCriterion (4ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
sqr#(s(X)) | → | add#(sqr(X), dbl(X)) | | add#(s(X), Y) | → | add#(X, Y) |
dbl#(s(X)) | → | dbl#(X) | | half#(s(s(X))) | → | half#(X) |
terms#(N) | → | sqr#(N) | | sqr#(s(X)) | → | dbl#(X) |
sqr#(s(X)) | → | sqr#(X) |
Rewrite Rules
terms(N) | → | cons(recip(sqr(N))) | | sqr(0) | → | 0 |
sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
first(s(X), cons(Y)) | → | cons(Y) | | half(0) | → | 0 |
half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
half(dbl(X)) | → | X |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, half, dbl, recip, add, first, cons, nil
Strategy
The following SCCs where found
add#(s(X), Y) → add#(X, Y) |
half#(s(s(X))) → half#(X) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
add#(s(X), Y) | → | add#(X, Y) |
Rewrite Rules
terms(N) | → | cons(recip(sqr(N))) | | sqr(0) | → | 0 |
sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
first(s(X), cons(Y)) | → | cons(Y) | | half(0) | → | 0 |
half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
half(dbl(X)) | → | X |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, half, dbl, recip, add, first, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
add#(s(X), Y) | → | add#(X, Y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
half#(s(s(X))) | → | half#(X) |
Rewrite Rules
terms(N) | → | cons(recip(sqr(N))) | | sqr(0) | → | 0 |
sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
first(s(X), cons(Y)) | → | cons(Y) | | half(0) | → | 0 |
half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
half(dbl(X)) | → | X |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, half, dbl, recip, add, first, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
half#(s(s(X))) | → | half#(X) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
terms(N) | → | cons(recip(sqr(N))) | | sqr(0) | → | 0 |
sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
first(s(X), cons(Y)) | → | cons(Y) | | half(0) | → | 0 |
half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
half(dbl(X)) | → | X |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, half, dbl, recip, add, first, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
terms(N) | → | cons(recip(sqr(N))) | | sqr(0) | → | 0 |
sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
first(s(X), cons(Y)) | → | cons(Y) | | half(0) | → | 0 |
half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
half(dbl(X)) | → | X |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, half, dbl, recip, add, first, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: