Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z092)
The rewrite relation of the following TRS is considered.
|
q0(0(x1)) |
→ |
0'(q1(x1)) |
(1) |
|
q1(0(x1)) |
→ |
0(q1(x1)) |
(2) |
|
q1(1'(x1)) |
→ |
1'(q1(x1)) |
(3) |
|
0(q1(1(x1))) |
→ |
q2(0(1'(x1))) |
(4) |
|
0'(q1(1(x1))) |
→ |
q2(0'(1'(x1))) |
(5) |
|
1'(q1(1(x1))) |
→ |
q2(1'(1'(x1))) |
(6) |
|
0(q2(0(x1))) |
→ |
q2(0(0(x1))) |
(7) |
|
0'(q2(0(x1))) |
→ |
q2(0'(0(x1))) |
(8) |
|
1'(q2(0(x1))) |
→ |
q2(1'(0(x1))) |
(9) |
|
0(q2(1'(x1))) |
→ |
q2(0(1'(x1))) |
(10) |
|
0'(q2(1'(x1))) |
→ |
q2(0'(1'(x1))) |
(11) |
|
1'(q2(1'(x1))) |
→ |
q2(1'(1'(x1))) |
(12) |
|
q2(0'(x1)) |
→ |
0'(q0(x1)) |
(13) |
|
q0(1'(x1)) |
→ |
1'(q3(x1)) |
(14) |
|
q3(1'(x1)) |
→ |
1'(q3(x1)) |
(15) |
|
q3(b(x1)) |
→ |
b(q4(x1)) |
(16) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by matchbox @ termCOMP 2023)
1 Rule Removal
Using the
matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
| [q1(x1)] |
= |
x1 +
|
| [q0(x1)] |
= |
x1 +
|
| [q3(x1)] |
= |
x1 +
|
| [q2(x1)] |
= |
x1 +
|
| [q4(x1)] |
= |
x1 +
|
| [0'(x1)] |
= |
x1 +
|
| [1'(x1)] |
= |
x1 +
|
| [1(x1)] |
= |
x1 +
|
| [0(x1)] |
= |
x1 +
|
| [b(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
|
q0(0(x1)) |
→ |
0'(q1(x1)) |
(1) |
|
0(q1(1(x1))) |
→ |
q2(0(1'(x1))) |
(4) |
|
0'(q1(1(x1))) |
→ |
q2(0'(1'(x1))) |
(5) |
|
1'(q1(1(x1))) |
→ |
q2(1'(1'(x1))) |
(6) |
|
q2(0'(x1)) |
→ |
0'(q0(x1)) |
(13) |
|
q0(1'(x1)) |
→ |
1'(q3(x1)) |
(14) |
|
q3(b(x1)) |
→ |
b(q4(x1)) |
(16) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
q1#(1'(x1)) |
→ |
q1#(x1) |
(17) |
|
q1#(1'(x1)) |
→ |
1'#(q1(x1)) |
(18) |
|
q1#(0(x1)) |
→ |
q1#(x1) |
(19) |
|
q1#(0(x1)) |
→ |
0#(q1(x1)) |
(20) |
|
q3#(1'(x1)) |
→ |
q3#(x1) |
(21) |
|
q3#(1'(x1)) |
→ |
1'#(q3(x1)) |
(22) |
|
0'#(q2(1'(x1))) |
→ |
0'#(1'(x1)) |
(23) |
|
0'#(q2(0(x1))) |
→ |
0'#(0(x1)) |
(24) |
|
1'#(q2(1'(x1))) |
→ |
1'#(1'(x1)) |
(25) |
|
1'#(q2(0(x1))) |
→ |
1'#(0(x1)) |
(26) |
|
0#(q2(1'(x1))) |
→ |
0#(1'(x1)) |
(27) |
|
0#(q2(0(x1))) |
→ |
0#(0(x1)) |
(28) |
1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
| [q1(x1)] |
= |
x1 +
|
| [q3(x1)] |
= |
x1 +
|
| [q2(x1)] |
= |
x1 +
|
| [0'(x1)] |
= |
x1 +
|
| [1'(x1)] |
= |
x1 +
|
| [0(x1)] |
= |
x1 +
|
| [q1#(x1)] |
= |
x1 +
|
| [q3#(x1)] |
= |
x1 +
|
| [0'#(x1)] |
= |
x1 +
|
| [1'#(x1)] |
= |
x1 +
|
| [0#(x1)] |
= |
x1 +
|
together with the usable
rules
|
q1(0(x1)) |
→ |
0(q1(x1)) |
(2) |
|
q1(1'(x1)) |
→ |
1'(q1(x1)) |
(3) |
|
0(q2(0(x1))) |
→ |
q2(0(0(x1))) |
(7) |
|
0'(q2(0(x1))) |
→ |
q2(0'(0(x1))) |
(8) |
|
1'(q2(0(x1))) |
→ |
q2(1'(0(x1))) |
(9) |
|
0(q2(1'(x1))) |
→ |
q2(0(1'(x1))) |
(10) |
|
0'(q2(1'(x1))) |
→ |
q2(0'(1'(x1))) |
(11) |
|
1'(q2(1'(x1))) |
→ |
q2(1'(1'(x1))) |
(12) |
|
q3(1'(x1)) |
→ |
1'(q3(x1)) |
(15) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
|
q1#(1'(x1)) |
→ |
q1#(x1) |
(17) |
|
q1#(1'(x1)) |
→ |
1'#(q1(x1)) |
(18) |
|
q1#(0(x1)) |
→ |
q1#(x1) |
(19) |
|
q1#(0(x1)) |
→ |
0#(q1(x1)) |
(20) |
|
q3#(1'(x1)) |
→ |
q3#(x1) |
(21) |
|
q3#(1'(x1)) |
→ |
1'#(q3(x1)) |
(22) |
|
0'#(q2(1'(x1))) |
→ |
0'#(1'(x1)) |
(23) |
|
0'#(q2(0(x1))) |
→ |
0'#(0(x1)) |
(24) |
|
1'#(q2(1'(x1))) |
→ |
1'#(1'(x1)) |
(25) |
|
1'#(q2(0(x1))) |
→ |
1'#(0(x1)) |
(26) |
|
0#(q2(1'(x1))) |
→ |
0#(1'(x1)) |
(27) |
|
0#(q2(0(x1))) |
→ |
0#(0(x1)) |
(28) |
and
no rules
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.