Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
del(x . (y . z)) |
→ f(x = y,x,y,z) |
| 2: |
|
f(true,x,y,z) |
→ del(y . z) |
| 3: |
|
f(false,x,y,z) |
→ x . del(y . z) |
| 4: |
|
nil = nil |
→ true |
| 5: |
|
(x . y) = nil |
→ false |
| 6: |
|
nil = (y . z) |
→ false |
| 7: |
|
(x . y) = (u . v) |
→ and(x = u,y = v) |
|
There are 6 dependency pairs:
|
| 8: |
|
DEL(x . (y . z)) |
→ F(x = y,x,y,z) |
| 9: |
|
DEL(x . (y . z)) |
→ x =# y |
| 10: |
|
F(true,x,y,z) |
→ DEL(y . z) |
| 11: |
|
F(false,x,y,z) |
→ DEL(y . z) |
| 12: |
|
(x . y) =# (u . v) |
→ x =# u |
| 13: |
|
(x . y) =# (u . v) |
→ y =# v |
|
The approximated dependency graph contains one SCC:
{8,10,11}.
-
Consider the SCC {8,10,11}.
The usable rules are {4-7}.
By taking the AF π with
π(and) = π(DEL) = 1,
π(=) = 2,
π(.) = [2]
and π(F) = [4] together with
the lexicographic path order with
precedence F ≻ false,
nil ≻ false,
nil ≻ true,
v ≻ u
and F ≈ .,
the rules in {10,11}
are weakly decreasing and
the rules in {4-8}
are strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.30 seconds)
--- May 4, 2006