Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
ack(0,y) |
→ s(y) |
| 2: |
|
ack(s(x),0) |
→ ack(x,s(0)) |
| 3: |
|
ack(s(x),s(y)) |
→ ack(x,ack(s(x),y)) |
|
There are 3 dependency pairs:
|
| 4: |
|
ACK(s(x),0) |
→ ACK(x,s(0)) |
| 5: |
|
ACK(s(x),s(y)) |
→ ACK(x,ack(s(x),y)) |
| 6: |
|
ACK(s(x),s(y)) |
→ ACK(s(x),y) |
|
The approximated dependency graph contains one SCC:
{4-6}.
-
Consider the SCC {4-6}.
By taking the AF π with
π(ACK) = 1 together with
the lexicographic path order with
precedence ack ≻ s,
rule 6
is weakly decreasing and
the rules in {1-5}
are strictly decreasing.
There is one new SCC.
-
Consider the SCC {6}.
There are no usable rules.
By taking the AF π with
π(ACK) = 2 together with
the lexicographic path order with
empty precedence,
rule 6
is strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.02 seconds)
--- May 4, 2006