Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
plus(0,Y) |
→ Y |
| 2: |
|
plus(s(X),Y) |
→ s(plus(X,Y)) |
| 3: |
|
min(X,0) |
→ X |
| 4: |
|
min(s(X),s(Y)) |
→ min(X,Y) |
| 5: |
|
min(min(X,Y),Z) |
→ min(X,plus(Y,Z)) |
| 6: |
|
quot(0,s(Y)) |
→ 0 |
| 7: |
|
quot(s(X),s(Y)) |
→ s(quot(min(X,Y),s(Y))) |
|
There are 6 dependency pairs:
|
| 8: |
|
PLUS(s(X),Y) |
→ PLUS(X,Y) |
| 9: |
|
MIN(s(X),s(Y)) |
→ MIN(X,Y) |
| 10: |
|
MIN(min(X,Y),Z) |
→ MIN(X,plus(Y,Z)) |
| 11: |
|
MIN(min(X,Y),Z) |
→ PLUS(Y,Z) |
| 12: |
|
QUOT(s(X),s(Y)) |
→ QUOT(min(X,Y),s(Y)) |
| 13: |
|
QUOT(s(X),s(Y)) |
→ MIN(X,Y) |
|
The approximated dependency graph contains 3 SCCs:
{8},
{9,10}
and {12}.
-
Consider the SCC {8}.
There are no usable rules.
By taking the AF π with
π(PLUS) = 1 together with
the lexicographic path order with
empty precedence,
rule 8
is strictly decreasing.
-
Consider the SCC {9,10}.
The usable rules are {1,2}.
By taking the AF π with
π(MIN) = π(s) = 1,
π(plus) = 2
and π(min) = [1] together with
the lexicographic path order with
empty precedence,
the rules in {1,2,9}
are weakly decreasing and
rule 10
is strictly decreasing.
There is one new SCC.
-
Consider the SCC {9}.
There are no usable rules.
By taking the AF π with
π(MIN) = 1 together with
the lexicographic path order with
empty precedence,
rule 9
is strictly decreasing.
-
Consider the SCC {12}.
The usable rules are {1-5}.
By taking the AF π with
π(min) = π(QUOT) = 1 together with
the lexicographic path order with
precedence plus ≻ s,
the rules in {3,5}
are weakly decreasing and
the rules in {1,2,4,12}
are strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.03 seconds)
--- May 4, 2006