Certification Problem
Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex1_Zan97_iGM)
The rewrite relation of the following TRS is considered.
|
active(g(X)) |
→ |
mark(h(X)) |
(1) |
|
active(c) |
→ |
mark(d) |
(2) |
|
active(h(d)) |
→ |
mark(g(c)) |
(3) |
|
mark(g(X)) |
→ |
active(g(X)) |
(4) |
|
mark(h(X)) |
→ |
active(h(X)) |
(5) |
|
mark(c) |
→ |
active(c) |
(6) |
|
mark(d) |
→ |
active(d) |
(7) |
|
g(mark(X)) |
→ |
g(X) |
(8) |
|
g(active(X)) |
→ |
g(X) |
(9) |
|
h(mark(X)) |
→ |
h(X) |
(10) |
|
h(active(X)) |
→ |
h(X) |
(11) |
The evaluation strategy is innermost.Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Constant to Unary
Every constant is turned into a unary function symbol to obtain the TRS
|
active(g(X)) |
→ |
mark(h(X)) |
(1) |
|
active(c'(x)) |
→ |
mark(d'(x)) |
(12) |
|
active(h(d'(x))) |
→ |
mark(g(c'(x))) |
(13) |
|
mark(g(X)) |
→ |
active(g(X)) |
(4) |
|
mark(h(X)) |
→ |
active(h(X)) |
(5) |
|
mark(c'(x)) |
→ |
active(c'(x)) |
(14) |
|
mark(d'(x)) |
→ |
active(d'(x)) |
(15) |
|
g(mark(X)) |
→ |
g(X) |
(8) |
|
g(active(X)) |
→ |
g(X) |
(9) |
|
h(mark(X)) |
→ |
h(X) |
(10) |
|
h(active(X)) |
→ |
h(X) |
(11) |
1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
g(active(X)) |
→ |
h(mark(X)) |
(16) |
|
c'(active(x)) |
→ |
d'(mark(x)) |
(17) |
|
d'(h(active(x))) |
→ |
c'(g(mark(x))) |
(18) |
|
g(mark(X)) |
→ |
g(active(X)) |
(19) |
|
h(mark(X)) |
→ |
h(active(X)) |
(20) |
|
c'(mark(x)) |
→ |
c'(active(x)) |
(21) |
|
d'(mark(x)) |
→ |
d'(active(x)) |
(22) |
|
mark(g(X)) |
→ |
g(X) |
(23) |
|
active(g(X)) |
→ |
g(X) |
(24) |
|
mark(h(X)) |
→ |
h(X) |
(25) |
|
active(h(X)) |
→ |
h(X) |
(26) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [g(x1)] |
= |
1 · x1
|
| [active(x1)] |
= |
1 · x1 + 1 |
| [h(x1)] |
= |
1 · x1
|
| [mark(x1)] |
= |
1 · x1 + 1 |
| [c'(x1)] |
= |
1 · x1
|
| [d'(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
mark(g(X)) |
→ |
g(X) |
(23) |
|
active(g(X)) |
→ |
g(X) |
(24) |
|
mark(h(X)) |
→ |
h(X) |
(25) |
|
active(h(X)) |
→ |
h(X) |
(26) |
1.1.1.1 Switch to Innermost Termination
The TRS is overlay and locally confluent:
10Hence, it suffices to show innermost termination in the following.
1.1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
g#(active(X)) |
→ |
h#(mark(X)) |
(27) |
|
c'#(active(x)) |
→ |
d'#(mark(x)) |
(28) |
|
d'#(h(active(x))) |
→ |
c'#(g(mark(x))) |
(29) |
|
d'#(h(active(x))) |
→ |
g#(mark(x)) |
(30) |
|
g#(mark(X)) |
→ |
g#(active(X)) |
(31) |
|
h#(mark(X)) |
→ |
h#(active(X)) |
(32) |
|
c'#(mark(x)) |
→ |
c'#(active(x)) |
(33) |
|
d'#(mark(x)) |
→ |
d'#(active(x)) |
(34) |
1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.