Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
ackin(0,X) |
→ ackout(s(X)) |
| 2: |
|
ackin(s(X),0) |
→ u11(ackin(X,s(0))) |
| 3: |
|
u11(ackout(X)) |
→ ackout(X) |
| 4: |
|
ackin(s(X),s(Y)) |
→ u21(ackin(s(X),Y),X) |
| 5: |
|
u21(ackout(X),Y) |
→ u22(ackin(Y,X)) |
| 6: |
|
u22(ackout(X)) |
→ ackout(X) |
|
There are 6 dependency pairs:
|
| 7: |
|
ACKIN(s(X),0) |
→ U11(ackin(X,s(0))) |
| 8: |
|
ACKIN(s(X),0) |
→ ACKIN(X,s(0)) |
| 9: |
|
ACKIN(s(X),s(Y)) |
→ U21(ackin(s(X),Y),X) |
| 10: |
|
ACKIN(s(X),s(Y)) |
→ ACKIN(s(X),Y) |
| 11: |
|
U21(ackout(X),Y) |
→ U22(ackin(Y,X)) |
| 12: |
|
U21(ackout(X),Y) |
→ ACKIN(Y,X) |
|
The approximated dependency graph contains one SCC:
{8-10,12}.
-
Consider the SCC {8-10,12}.
By taking the AF π with
π(ackin) = π(ACKIN) = π(u11) = π(u21) = 1,
π(U21) = 2
and π(ackout) = π(u22) = [ ] together with
the lexicographic path order with
precedence 0 ≻ u22
and u22 ≈ ackout,
the rules in {3-6,10,12}
are weakly decreasing and
the rules in {1,2,8,9}
are strictly decreasing.
There is one new SCC.
-
Consider the SCC {10}.
There are no usable rules.
By taking the AF π with
π(ACKIN) = 2 together with
the lexicographic path order with
empty precedence,
rule 10
is strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.12 seconds)
--- May 4, 2006