Certification Problem
Input (TPDB TRS_Standard/SK90/2.27)
The rewrite relation of the following TRS is considered.
|
fib(0) |
→ |
0 |
(1) |
|
fib(s(0)) |
→ |
s(0) |
(2) |
|
fib(s(s(0))) |
→ |
s(0) |
(3) |
|
fib(s(s(x))) |
→ |
sp(g(x)) |
(4) |
|
g(0) |
→ |
pair(s(0),0) |
(5) |
|
g(s(0)) |
→ |
pair(s(0),s(0)) |
(6) |
|
g(s(x)) |
→ |
np(g(x)) |
(7) |
|
sp(pair(x,y)) |
→ |
+(x,y) |
(8) |
|
np(pair(x,y)) |
→ |
pair(+(x,y),x) |
(9) |
|
+(x,0) |
→ |
x |
(10) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(11) |
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.
|
+#(x,s(y)) |
→ |
+#(x,y) |
(12) |
|
fib#(s(s(x))) |
→ |
sp#(g(x)) |
(13) |
|
fib#(s(s(x))) |
→ |
g#(x) |
(14) |
|
sp#(pair(x,y)) |
→ |
+#(x,y) |
(15) |
|
g#(s(x)) |
→ |
np#(g(x)) |
(16) |
|
np#(pair(x,y)) |
→ |
+#(x,y) |
(17) |
|
g#(s(x)) |
→ |
g#(x) |
(18) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.