Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
times(x,0) |
→ 0 |
| 2: |
|
times(x,s(y)) |
→ plus(times(x,y),x) |
| 3: |
|
plus(x,0) |
→ x |
| 4: |
|
plus(0,x) |
→ x |
| 5: |
|
plus(x,s(y)) |
→ s(plus(x,y)) |
| 6: |
|
plus(s(x),y) |
→ s(plus(x,y)) |
|
There are 4 dependency pairs:
|
| 7: |
|
TIMES(x,s(y)) |
→ PLUS(times(x,y),x) |
| 8: |
|
TIMES(x,s(y)) |
→ TIMES(x,y) |
| 9: |
|
PLUS(x,s(y)) |
→ PLUS(x,y) |
| 10: |
|
PLUS(s(x),y) |
→ PLUS(x,y) |
|
The approximated dependency graph contains 2 SCCs:
{9,10}
and {8}.
-
Consider the SCC {9,10}.
There are no usable rules.
By taking the AF π with
π(PLUS) = 1 together with
the lexicographic path order with
empty precedence,
rule 9
is weakly decreasing and
rule 10
is strictly decreasing.
There is one new SCC.
-
Consider the SCC {9}.
By taking the AF π with
π(PLUS) = 2 together with
the lexicographic path order with
empty precedence,
rule 9
is strictly decreasing.
-
Consider the SCC {8}.
There are no usable rules.
By taking the AF π with
π(TIMES) = 2 together with
the lexicographic path order with
empty precedence,
rule 8
is strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.01 seconds)
--- May 3, 2006