The rewrite relation of the following TRS is considered.
active(f(X,X)) | → | mark(f(a,b)) | (1) |
active(b) | → | mark(a) | (2) |
mark(f(X1,X2)) | → | active(f(mark(X1),X2)) | (3) |
mark(a) | → | active(a) | (4) |
mark(b) | → | active(b) | (5) |
f(mark(X1),X2) | → | f(X1,X2) | (6) |
f(X1,mark(X2)) | → | f(X1,X2) | (7) |
f(active(X1),X2) | → | f(X1,X2) | (8) |
f(X1,active(X2)) | → | f(X1,X2) | (9) |
active#(f(X,X)) | → | mark#(f(a,b)) | (10) |
active#(f(X,X)) | → | f#(a,b) | (11) |
active#(b) | → | mark#(a) | (12) |
mark#(f(X1,X2)) | → | active#(f(mark(X1),X2)) | (13) |
mark#(f(X1,X2)) | → | f#(mark(X1),X2) | (14) |
mark#(f(X1,X2)) | → | mark#(X1) | (15) |
mark#(a) | → | active#(a) | (16) |
mark#(b) | → | active#(b) | (17) |
f#(mark(X1),X2) | → | f#(X1,X2) | (18) |
f#(X1,mark(X2)) | → | f#(X1,X2) | (19) |
f#(active(X1),X2) | → | f#(X1,X2) | (20) |
f#(X1,active(X2)) | → | f#(X1,X2) | (21) |
The dependency pairs are split into 2 components.
mark#(f(X1,X2)) | → | active#(f(mark(X1),X2)) | (13) |
active#(f(X,X)) | → | mark#(f(a,b)) | (10) |
mark#(f(X1,X2)) | → | mark#(X1) | (15) |
[active(x1)] | = | 1 · x1 |
[f(x1, x2)] | = | 1 + 2 · x1 + 2 · x2 |
[mark(x1)] | = | 1 · x1 |
[a] | = | 0 |
[b] | = | 0 |
[mark#(x1)] | = | 2 + 2 · x1 |
[active#(x1)] | = | 2 + 2 · x1 |
mark#(f(X1,X2)) | → | mark#(X1) | (15) |
mark#(f(a,b)) | → | active#(f(mark(a),b)) | (22) |
We restrict the rewrite rules to the following usable rules of the DP problem.
mark(a) | → | active(a) | (4) |
f(X1,mark(X2)) | → | f(X1,X2) | (7) |
f(mark(X1),X2) | → | f(X1,X2) | (6) |
f(active(X1),X2) | → | f(X1,X2) | (8) |
f(X1,active(X2)) | → | f(X1,X2) | (9) |
→ |
We restrict the rewrite rules to the following usable rules of the DP problem.
f(X1,mark(X2)) | → | f(X1,X2) | (7) |
f(mark(X1),X2) | → | f(X1,X2) | (6) |
f(active(X1),X2) | → | f(X1,X2) | (8) |
f(X1,active(X2)) | → | f(X1,X2) | (9) |
[f(x1, x2)] | = | 1 · x1 + 2 · x2 |
[mark(x1)] | = | 2 · x1 |
[active(x1)] | = | 1 · x1 |
[active#(x1)] | = | 1 · x1 |
[mark#(x1)] | = | 1 · x1 |
[a] | = | 0 |
[b] | = | 0 |
f(X1,mark(X2)) | → | f(X1,X2) | (7) |
f(mark(X1),X2) | → | f(X1,X2) | (6) |
f(active(X1),X2) | → | f(X1,X2) | (8) |
f(X1,active(X2)) | → | f(X1,X2) | (9) |
f(X1,mark(X2)) | → | f(X1,X2) | (7) |
f(mark(X1),X2) | → | f(X1,X2) | (6) |
We restrict the innermost strategy to the following left hand sides.
active(f(x0,x0)) |
active(b) |
f(mark(x0),x1) |
f(x0,mark(x1)) |
f(active(x0),x1) |
f(x0,active(x1)) |
mark#(f(a,b)) | → | active#(f(a,b)) | (24) |
The dependency pairs are split into 0 components.
f#(X1,mark(X2)) | → | f#(X1,X2) | (19) |
f#(mark(X1),X2) | → | f#(X1,X2) | (18) |
f#(active(X1),X2) | → | f#(X1,X2) | (20) |
f#(X1,active(X2)) | → | f#(X1,X2) | (21) |
We restrict the rewrite rules to the following usable rules of the DP problem.
There are no rules.
We restrict the innermost strategy to the following left hand sides.
active(f(x0,x0)) |
active(b) |
mark(f(x0,x1)) |
mark(a) |
mark(b) |
Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.
f#(X1,mark(X2)) | → | f#(X1,X2) | (19) |
1 | ≥ | 1 | |
2 | > | 2 | |
f#(mark(X1),X2) | → | f#(X1,X2) | (18) |
1 | > | 1 | |
2 | ≥ | 2 | |
f#(active(X1),X2) | → | f#(X1,X2) | (20) |
1 | > | 1 | |
2 | ≥ | 2 | |
f#(X1,active(X2)) | → | f#(X1,X2) | (21) |
1 | ≥ | 1 | |
2 | > | 2 |
As there is no critical graph in the transitive closure, there are no infinite chains.