Certification Problem
Input (TPDB TRS_Standard/Various_04/12)
The rewrite relation of the following TRS is considered.
|
O(0) |
→ |
0 |
(1) |
|
+(0,x) |
→ |
x |
(2) |
|
+(x,0) |
→ |
x |
(3) |
|
+(O(x),O(y)) |
→ |
O(+(x,y)) |
(4) |
|
+(O(x),I(y)) |
→ |
I(+(x,y)) |
(5) |
|
+(I(x),O(y)) |
→ |
I(+(x,y)) |
(6) |
|
+(I(x),I(y)) |
→ |
O(+(+(x,y),I(0))) |
(7) |
|
*(0,x) |
→ |
0 |
(8) |
|
*(x,0) |
→ |
0 |
(9) |
|
*(O(x),y) |
→ |
O(*(x,y)) |
(10) |
|
*(I(x),y) |
→ |
+(O(*(x,y)),y) |
(11) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
+#(O(x),O(y)) |
→ |
O#(+(x,y)) |
(12) |
|
+#(O(x),O(y)) |
→ |
+#(x,y) |
(13) |
|
+#(O(x),I(y)) |
→ |
+#(x,y) |
(14) |
|
+#(I(x),O(y)) |
→ |
+#(x,y) |
(15) |
|
+#(I(x),I(y)) |
→ |
O#(+(+(x,y),I(0))) |
(16) |
|
+#(I(x),I(y)) |
→ |
+#(+(x,y),I(0)) |
(17) |
|
+#(I(x),I(y)) |
→ |
+#(x,y) |
(18) |
|
*#(O(x),y) |
→ |
O#(*(x,y)) |
(19) |
|
*#(O(x),y) |
→ |
*#(x,y) |
(20) |
|
*#(I(x),y) |
→ |
+#(O(*(x,y)),y) |
(21) |
|
*#(I(x),y) |
→ |
O#(*(x,y)) |
(22) |
|
*#(I(x),y) |
→ |
*#(x,y) |
(23) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
*#(I(x),y) |
→ |
*#(x,y) |
(23) |
|
*#(O(x),y) |
→ |
*#(x,y) |
(20) |
1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [I(x1)] |
= |
1 · x1
|
| [O(x1)] |
= |
1 · x1
|
| [*#(x1, x2)] |
= |
1 · x1 + 1 · x2
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
rule
could be deleted.
1.1.1.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
*#(I(x),y) |
→ |
*#(x,y) |
(23) |
|
|
| 1 |
> |
1 |
| 2 |
≥ |
2 |
|
*#(O(x),y) |
→ |
*#(x,y) |
(20) |
|
|
| 1 |
> |
1 |
| 2 |
≥ |
2 |
As there is no critical graph in the transitive closure, there are no infinite chains.
-
The
2nd
component contains the
pair
|
+#(O(x),I(y)) |
→ |
+#(x,y) |
(14) |
|
+#(O(x),O(y)) |
→ |
+#(x,y) |
(13) |
|
+#(I(x),O(y)) |
→ |
+#(x,y) |
(15) |
|
+#(I(x),I(y)) |
→ |
+#(+(x,y),I(0)) |
(17) |
|
+#(I(x),I(y)) |
→ |
+#(x,y) |
(18) |
1.1.2 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [+(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [0] |
= |
0 |
| [O(x1)] |
= |
1 · x1
|
| [I(x1)] |
= |
1 · x1
|
| [+#(x1, x2)] |
= |
1 · x1 + 1 · x2
|
together with the usable
rules
|
+(0,x) |
→ |
x |
(2) |
|
+(x,0) |
→ |
x |
(3) |
|
+(O(x),O(y)) |
→ |
O(+(x,y)) |
(4) |
|
+(O(x),I(y)) |
→ |
I(+(x,y)) |
(5) |
|
+(I(x),O(y)) |
→ |
I(+(x,y)) |
(6) |
|
+(I(x),I(y)) |
→ |
O(+(+(x,y),I(0))) |
(7) |
|
O(0) |
→ |
0 |
(1) |
(w.r.t. the implicit argument filter of the reduction pair),
the
rule
could be deleted.
1.1.2.1 Monotonic Reduction Pair Processor
Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(0) |
= |
4 |
|
weight(0) |
= |
2 |
|
|
|
| prec(O) |
= |
0 |
|
weight(O) |
= |
1 |
|
|
|
| prec(I) |
= |
1 |
|
weight(I) |
= |
3 |
|
|
|
| prec(+) |
= |
2 |
|
weight(+) |
= |
0 |
|
|
|
| prec(+#) |
= |
3 |
|
weight(+#) |
= |
0 |
|
|
|
the
pairs
|
+#(O(x),I(y)) |
→ |
+#(x,y) |
(14) |
|
+#(O(x),O(y)) |
→ |
+#(x,y) |
(13) |
|
+#(I(x),O(y)) |
→ |
+#(x,y) |
(15) |
|
+#(I(x),I(y)) |
→ |
+#(+(x,y),I(0)) |
(17) |
|
+#(I(x),I(y)) |
→ |
+#(x,y) |
(18) |
and
the
rules
|
+(0,x) |
→ |
x |
(2) |
|
+(x,0) |
→ |
x |
(3) |
|
+(O(x),O(y)) |
→ |
O(+(x,y)) |
(4) |
|
+(O(x),I(y)) |
→ |
I(+(x,y)) |
(5) |
|
+(I(x),O(y)) |
→ |
I(+(x,y)) |
(6) |
|
+(I(x),I(y)) |
→ |
O(+(+(x,y),I(0))) |
(7) |
|
O(0) |
→ |
0 |
(1) |
could be deleted.
1.1.2.1.1 P is empty
There are no pairs anymore.