Certification Problem
Input (TPDB SRS_Standard/Trafo_06/dup06)
The rewrite relation of the following TRS is considered.
a(a(b(b(x1)))) |
→ |
b(b(c(c(a(a(x1)))))) |
(1) |
b(b(c(c(x1)))) |
→ |
c(c(b(b(b(b(x1)))))) |
(2) |
b(b(a(a(x1)))) |
→ |
a(a(c(c(b(b(x1)))))) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
[a(x1)] |
= |
· x1 +
|
[b(x1)] |
= |
· x1 +
|
[c(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
a(a(b(b(x1)))) |
→ |
b(b(c(c(a(a(x1)))))) |
(1) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
b#(b(c(c(x1)))) |
→ |
b#(x1) |
(4) |
b#(b(c(c(x1)))) |
→ |
b#(b(x1)) |
(5) |
b#(b(c(c(x1)))) |
→ |
b#(b(b(x1))) |
(6) |
b#(b(c(c(x1)))) |
→ |
b#(b(b(b(x1)))) |
(7) |
b#(b(a(a(x1)))) |
→ |
b#(x1) |
(8) |
b#(b(a(a(x1)))) |
→ |
b#(b(x1)) |
(9) |
1.1.1 Subterm Criterion Processor
We use the projection to multisets
π(b#)
|
= |
{
1, 1
}
|
π(c)
|
= |
{
1
}
|
π(a)
|
= |
{
1, 1
}
|
π(b)
|
= |
{
1
}
|
to remove the pairs:
b#(b(a(a(x1)))) |
→ |
b#(x1) |
(8) |
b#(b(a(a(x1)))) |
→ |
b#(b(x1)) |
(9) |
1.1.1.1 Reduction Pair Processor with Usable Rules
Using the
prec(b#) |
= |
0 |
|
stat(b#) |
= |
lex
|
prec(c) |
= |
0 |
|
stat(c) |
= |
lex
|
prec(a) |
= |
0 |
|
stat(a) |
= |
lex
|
prec(b) |
= |
0 |
|
stat(b) |
= |
lex
|
π(b#) |
= |
1 |
π(c) |
= |
[1] |
π(a) |
= |
[] |
π(b) |
= |
1 |
together with the usable
rules
b(b(c(c(x1)))) |
→ |
c(c(b(b(b(b(x1)))))) |
(2) |
b(b(a(a(x1)))) |
→ |
a(a(c(c(b(b(x1)))))) |
(3) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
b#(b(c(c(x1)))) |
→ |
b#(x1) |
(4) |
b#(b(c(c(x1)))) |
→ |
b#(b(x1)) |
(5) |
b#(b(c(c(x1)))) |
→ |
b#(b(b(x1))) |
(6) |
b#(b(c(c(x1)))) |
→ |
b#(b(b(b(x1)))) |
(7) |
could be deleted.
1.1.1.1.1 P is empty
There are no pairs anymore.