Certification Problem
Input (TPDB TRS_Standard/SK90/2.26)
The rewrite relation of the following TRS is considered.
|
f(0) |
→ |
0 |
(1) |
|
f(s(0)) |
→ |
s(0) |
(2) |
|
f(s(s(x))) |
→ |
p(h(g(x))) |
(3) |
|
g(0) |
→ |
pair(s(0),s(0)) |
(4) |
|
g(s(x)) |
→ |
h(g(x)) |
(5) |
|
h(x) |
→ |
pair(+(p(x),q(x)),p(x)) |
(6) |
|
p(pair(x,y)) |
→ |
x |
(7) |
|
q(pair(x,y)) |
→ |
y |
(8) |
|
+(x,0) |
→ |
x |
(9) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(10) |
|
f(s(s(x))) |
→ |
+(p(g(x)),q(g(x))) |
(11) |
|
g(s(x)) |
→ |
pair(+(p(g(x)),q(g(x))),p(g(x))) |
(12) |
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.
|
h#(x) |
→ |
p#(x) |
(13) |
|
f#(s(s(x))) |
→ |
p#(g(x)) |
(14) |
|
g#(s(x)) |
→ |
q#(g(x)) |
(15) |
|
f#(s(s(x))) |
→ |
g#(x) |
(16) |
|
g#(s(x)) |
→ |
p#(g(x)) |
(17) |
|
g#(s(x)) |
→ |
g#(x) |
(18) |
|
g#(s(x)) |
→ |
g#(x) |
(18) |
|
f#(s(s(x))) |
→ |
p#(h(g(x))) |
(19) |
|
f#(s(s(x))) |
→ |
g#(x) |
(16) |
|
g#(s(x)) |
→ |
+#(p(g(x)),q(g(x))) |
(20) |
|
f#(s(s(x))) |
→ |
+#(p(g(x)),q(g(x))) |
(21) |
|
f#(s(s(x))) |
→ |
g#(x) |
(16) |
|
+#(x,s(y)) |
→ |
+#(x,y) |
(22) |
|
g#(s(x)) |
→ |
g#(x) |
(18) |
|
f#(s(s(x))) |
→ |
h#(g(x)) |
(23) |
|
g#(s(x)) |
→ |
g#(x) |
(18) |
|
h#(x) |
→ |
q#(x) |
(24) |
|
h#(x) |
→ |
+#(p(x),q(x)) |
(25) |
|
f#(s(s(x))) |
→ |
q#(g(x)) |
(26) |
|
g#(s(x)) |
→ |
p#(g(x)) |
(17) |
|
h#(x) |
→ |
p#(x) |
(13) |
|
g#(s(x)) |
→ |
h#(g(x)) |
(27) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.