The rewrite relation of the following TRS is considered.
| f(0) | → | cons(0,n__f(s(0))) | (1) |
| f(s(0)) | → | f(p(s(0))) | (2) |
| p(s(0)) | → | 0 | (3) |
| f(X) | → | n__f(X) | (4) |
| activate(n__f(X)) | → | f(X) | (5) |
| activate(X) | → | X | (6) |
| [f(x1)] | = | 1 + 2 · x1 |
| [0] | = | 1 |
| [cons(x1, x2)] | = | 1 · x1 + 1 · x2 |
| [n__f(x1)] | = | 1 · x1 |
| [s(x1)] | = | 2 · x1 |
| [p(x1)] | = | 1 · x1 |
| [activate(x1)] | = | 2 + 2 · x1 |
| p(s(0)) | → | 0 | (3) |
| f(X) | → | n__f(X) | (4) |
| activate(n__f(X)) | → | f(X) | (5) |
| activate(X) | → | X | (6) |
| prec(f) | = | 1 | stat(f) | = | mul | |
| prec(0) | = | 2 | stat(0) | = | mul | |
| prec(cons) | = | 0 | stat(cons) | = | mul |
| π(f) | = | [1] |
| π(0) | = | [] |
| π(cons) | = | [1,2] |
| π(n__f) | = | 1 |
| π(s) | = | 1 |
| π(p) | = | 1 |
| f(0) | → | cons(0,n__f(s(0))) | (1) |
The TRS is overlay and locally confluent:
10Hence, it suffices to show innermost termination in the following.
| f#(s(0)) | → | f#(p(s(0))) | (7) |
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.
There are no lhss.
final states:
{0, 1, 2, 3}
transitions:
| f#0(0) | → | 0 |
| f#0(1) | → | 0 |
| f#0(2) | → | 0 |
| f#0(3) | → | 0 |
| s0(0) | → | 1 |
| s0(1) | → | 1 |
| s0(2) | → | 1 |
| s0(3) | → | 1 |
| 00 | → | 2 |
| p0(0) | → | 3 |
| p0(1) | → | 3 |
| p0(2) | → | 3 |
| p0(3) | → | 3 |
| 01 | → | 6 |
| s1(6) | → | 5 |
| p1(5) | → | 4 |
| f#1(4) | → | 0 |