Certification Problem
Input (TPDB TRS_Standard/Secret_07_TRS/secret2)
The rewrite relation of the following TRS is considered.
|
a(h,h,h,x) |
→ |
s(x) |
(1) |
|
a(l,x,s(y),h) |
→ |
a(l,x,y,s(h)) |
(2) |
|
a(l,x,s(y),s(z)) |
→ |
a(l,x,y,a(l,x,s(y),z)) |
(3) |
|
a(l,s(x),h,z) |
→ |
a(l,x,z,z) |
(4) |
|
a(s(l),h,h,z) |
→ |
a(l,z,h,z) |
(5) |
|
+(x,h) |
→ |
x |
(6) |
|
+(h,x) |
→ |
x |
(7) |
|
+(s(x),s(y)) |
→ |
s(s(+(x,y))) |
(8) |
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(9) |
|
s(h) |
→ |
1 |
(10) |
|
*(h,x) |
→ |
h |
(11) |
|
*(x,h) |
→ |
h |
(12) |
|
*(s(x),s(y)) |
→ |
s(+(+(*(x,y),x),y)) |
(13) |
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#(l,x,s(y),h) |
→ |
s#(h) |
(14) |
|
*#(s(x),s(y)) |
→ |
*#(x,y) |
(15) |
|
+#(s(x),s(y)) |
→ |
s#(s(+(x,y))) |
(16) |
|
a#(s(l),h,h,z) |
→ |
a#(l,z,h,z) |
(17) |
|
a#(l,x,s(y),s(z)) |
→ |
a#(l,x,s(y),z) |
(18) |
|
a#(h,h,h,x) |
→ |
s#(x) |
(19) |
|
+#(s(x),s(y)) |
→ |
s#(+(x,y)) |
(20) |
|
+#(+(x,y),z) |
→ |
+#(x,+(y,z)) |
(21) |
|
a#(l,x,s(y),s(z)) |
→ |
a#(l,x,y,a(l,x,s(y),z)) |
(22) |
|
*#(s(x),s(y)) |
→ |
+#(*(x,y),x) |
(23) |
|
a#(l,s(x),h,z) |
→ |
a#(l,x,z,z) |
(24) |
|
*#(s(x),s(y)) |
→ |
s#(+(+(*(x,y),x),y)) |
(25) |
|
a#(l,x,s(y),h) |
→ |
a#(l,x,y,s(h)) |
(26) |
|
+#(+(x,y),z) |
→ |
+#(y,z) |
(27) |
|
+#(s(x),s(y)) |
→ |
+#(x,y) |
(28) |
|
*#(s(x),s(y)) |
→ |
+#(+(*(x,y),x),y) |
(29) |
1.1 Dependency Graph Processor
The dependency pairs are split into 3
components.
-
The
1st
component contains the
pair
|
*#(s(x),s(y)) |
→ |
*#(x,y) |
(15) |
1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [a(x1,...,x4)] |
=
|
0 |
| [h] |
=
|
0 |
| [1] |
=
|
0 |
| [s(x1)] |
=
|
x1 + 1 |
| [*#(x1, x2)] |
=
|
x1 + x2 + 0 |
| [s#(x1)] |
=
|
0 |
| [a#(x1,...,x4)] |
=
|
0 |
| [+(x1, x2)] |
=
|
0 |
| [+#(x1, x2)] |
=
|
0 |
| [*(x1, x2)] |
=
|
0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
|
*#(s(x),s(y)) |
→ |
*#(x,y) |
(15) |
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.
-
The
2nd
component contains the
pair
|
a#(l,x,s(y),h) |
→ |
a#(l,x,y,s(h)) |
(26) |
|
a#(l,s(x),h,z) |
→ |
a#(l,x,z,z) |
(24) |
|
a#(l,x,s(y),s(z)) |
→ |
a#(l,x,s(y),z) |
(18) |
|
a#(s(l),h,h,z) |
→ |
a#(l,z,h,z) |
(17) |
|
a#(l,x,s(y),s(z)) |
→ |
a#(l,x,y,a(l,x,s(y),z)) |
(22) |
1.1.2 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [a(x1,...,x4)] |
=
|
1 |
| [h] |
=
|
1 |
| [1] |
=
|
4 |
| [s(x1)] |
=
|
x1 + 2 |
| [*#(x1, x2)] |
=
|
0 |
| [s#(x1)] |
=
|
0 |
| [a#(x1,...,x4)] |
=
|
x1 + 0 |
| [+(x1, x2)] |
=
|
0 |
| [+#(x1, x2)] |
=
|
0 |
| [*(x1, x2)] |
=
|
0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
|
a#(s(l),h,h,z) |
→ |
a#(l,z,h,z) |
(17) |
could be deleted.
1.1.2.1 Dependency Graph Processor
The dependency pairs are split into 1
component.
-
The
3rd
component contains the
pair
|
+#(s(x),s(y)) |
→ |
+#(x,y) |
(28) |
|
+#(+(x,y),z) |
→ |
+#(y,z) |
(27) |
|
+#(+(x,y),z) |
→ |
+#(x,+(y,z)) |
(21) |
1.1.3 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [a(x1,...,x4)] |
=
|
x1 + x2 + x3 + x4 + 52869 |
| [h] |
=
|
1 |
| [1] |
=
|
52869 |
| [s(x1)] |
=
|
x1 + 52868 |
| [*#(x1, x2)] |
=
|
0 |
| [s#(x1)] |
=
|
0 |
| [a#(x1,...,x4)] |
=
|
x3 + 0 |
| [+(x1, x2)] |
=
|
x1 + x2 + 1 |
| [+#(x1, x2)] |
=
|
x1 + x2 + 0 |
| [*(x1, x2)] |
=
|
0 |
together with the usable
rules
|
+(s(x),s(y)) |
→ |
s(s(+(x,y))) |
(8) |
|
s(h) |
→ |
1 |
(10) |
|
+(h,x) |
→ |
x |
(7) |
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(9) |
|
+(x,h) |
→ |
x |
(6) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
|
+#(s(x),s(y)) |
→ |
+#(x,y) |
(28) |
|
+#(+(x,y),z) |
→ |
+#(y,z) |
(27) |
could be deleted.
1.1.3.1 Dependency Graph Processor
The dependency pairs are split into 1
component.