Certification Problem
Input (TPDB TRS_Standard/SK90/2.08)
The rewrite relation of the following TRS is considered.
|
+(a,b) |
→ |
+(b,a) |
(1) |
|
+(a,+(b,z)) |
→ |
+(b,+(a,z)) |
(2) |
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(3) |
|
f(a,y) |
→ |
a |
(4) |
|
f(b,y) |
→ |
b |
(5) |
|
f(+(x,y),z) |
→ |
+(f(x,z),f(y,z)) |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by NaTT @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
+#(a,+(b,z)) |
→ |
+#(a,z) |
(7) |
|
+#(+(x,y),z) |
→ |
+#(x,+(y,z)) |
(8) |
|
+#(+(x,y),z) |
→ |
+#(y,z) |
(9) |
|
f#(+(x,y),z) |
→ |
f#(y,z) |
(10) |
|
f#(+(x,y),z) |
→ |
+#(f(x,z),f(y,z)) |
(11) |
|
+#(a,+(b,z)) |
→ |
+#(b,+(a,z)) |
(12) |
|
+#(a,b) |
→ |
+#(b,a) |
(13) |
|
f#(+(x,y),z) |
→ |
f#(x,z) |
(14) |
1.1 Dependency Graph Processor
The dependency pairs are split into 3
components.
-
The
1st
component contains the
pair
|
f#(+(x,y),z) |
→ |
f#(x,z) |
(14) |
|
f#(+(x,y),z) |
→ |
f#(y,z) |
(10) |
1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [a] |
=
|
0 |
| [b] |
=
|
0 |
| [f(x1, x2)] |
=
|
0 |
| [f#(x1, x2)] |
=
|
x1 + 0 |
| [+(x1, x2)] |
=
|
x1 + x2 + 1 |
| [+#(x1, x2)] |
=
|
0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
|
f#(+(x,y),z) |
→ |
f#(x,z) |
(14) |
|
f#(+(x,y),z) |
→ |
f#(y,z) |
(10) |
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.
-
The
2nd
component contains the
pair
|
+#(+(x,y),z) |
→ |
+#(x,+(y,z)) |
(8) |
|
+#(+(x,y),z) |
→ |
+#(y,z) |
(9) |
1.1.2 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [a] |
=
|
1 |
| [b] |
=
|
23676 |
| [f(x1, x2)] |
=
|
0 |
| [f#(x1, x2)] |
=
|
0 |
| [+(x1, x2)] |
=
|
x1 + x2 + 1 |
| [+#(x1, x2)] |
=
|
x1 + 0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
|
+#(+(x,y),z) |
→ |
+#(x,+(y,z)) |
(8) |
|
+#(+(x,y),z) |
→ |
+#(y,z) |
(9) |
could be deleted.
1.1.2.1 Dependency Graph Processor
The dependency pairs are split into 0
components.
-
The
3rd
component contains the
pair
|
+#(a,+(b,z)) |
→ |
+#(a,z) |
(7) |
1.1.3 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [a] |
=
|
1 |
| [b] |
=
|
23676 |
| [f(x1, x2)] |
=
|
0 |
| [f#(x1, x2)] |
=
|
0 |
| [+(x1, x2)] |
=
|
x1 + x2 + 20653 |
| [+#(x1, x2)] |
=
|
x2 + 0 |
together with the usable
rules
|
+(a,b) |
→ |
+(b,a) |
(1) |
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(3) |
|
+(a,+(b,z)) |
→ |
+(b,+(a,z)) |
(2) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pair
|
+#(a,+(b,z)) |
→ |
+#(a,z) |
(7) |
could be deleted.
1.1.3.1 Dependency Graph Processor
The dependency pairs are split into 0
components.