YES
The TRS could be proven terminating. The proof took 72 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (15ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (3ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
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) | | 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) |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil
Strategy
The following SCCs where found
| add#(s(X), Y) → add#(X, Y) |
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) |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, 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
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) |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
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) |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: