Certification Problem
Input (TPDB SRS_Relative/Zantema_06_relative/rel02)
The relative rewrite relation R/S is considered where R is the following TRS
|
a(a(b(x1))) |
→ |
b(a(x1)) |
(1) |
|
c(b(x1)) |
→ |
b(a(b(x1))) |
(2) |
and S is the following TRS.
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
b(a(a(x1))) |
→ |
a(b(x1)) |
(4) |
|
b(c(x1)) |
→ |
b(a(b(x1))) |
(5) |
|
a(x1) |
→ |
a(c(a(x1))) |
(3) |
1.1 Closure Under Flat Contexts
Using the flat contexts
{b(☐), a(☐), c(☐)}
We obtain the transformed TRS
|
b(c(x1)) |
→ |
b(a(b(x1))) |
(5) |
|
b(b(a(a(x1)))) |
→ |
b(a(b(x1))) |
(6) |
|
a(b(a(a(x1)))) |
→ |
a(a(b(x1))) |
(7) |
|
c(b(a(a(x1)))) |
→ |
c(a(b(x1))) |
(8) |
|
a(x1) |
→ |
a(c(a(x1))) |
(3) |
1.1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
bc(cb(x1)) |
→ |
ba(ab(bb(x1))) |
(9) |
|
bc(cc(x1)) |
→ |
ba(ab(bc(x1))) |
(10) |
|
bc(ca(x1)) |
→ |
ba(ab(ba(x1))) |
(11) |
|
bb(ba(aa(ab(x1)))) |
→ |
ba(ab(bb(x1))) |
(12) |
|
bb(ba(aa(ac(x1)))) |
→ |
ba(ab(bc(x1))) |
(13) |
|
bb(ba(aa(aa(x1)))) |
→ |
ba(ab(ba(x1))) |
(14) |
|
ab(ba(aa(ab(x1)))) |
→ |
aa(ab(bb(x1))) |
(15) |
|
ab(ba(aa(ac(x1)))) |
→ |
aa(ab(bc(x1))) |
(16) |
|
ab(ba(aa(aa(x1)))) |
→ |
aa(ab(ba(x1))) |
(17) |
|
cb(ba(aa(ab(x1)))) |
→ |
ca(ab(bb(x1))) |
(18) |
|
cb(ba(aa(ac(x1)))) |
→ |
ca(ab(bc(x1))) |
(19) |
|
cb(ba(aa(aa(x1)))) |
→ |
ca(ab(ba(x1))) |
(20) |
|
ab(x1) |
→ |
ac(ca(ab(x1))) |
(21) |
|
ac(x1) |
→ |
ac(ca(ac(x1))) |
(22) |
|
aa(x1) |
→ |
ac(ca(aa(x1))) |
(23) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [bc(x1)] |
= |
1 · x1
|
| [cb(x1)] |
= |
1 + 1 · x1
|
| [ba(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 · x1
|
| [cc(x1)] |
= |
1 · x1
|
| [ca(x1)] |
= |
1 · x1
|
| [aa(x1)] |
= |
1 + 1 · x1
|
| [ac(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
bc(cb(x1)) |
→ |
ba(ab(bb(x1))) |
(9) |
|
bb(ba(aa(ab(x1)))) |
→ |
ba(ab(bb(x1))) |
(12) |
|
bb(ba(aa(ac(x1)))) |
→ |
ba(ab(bc(x1))) |
(13) |
|
bb(ba(aa(aa(x1)))) |
→ |
ba(ab(ba(x1))) |
(14) |
|
ab(ba(aa(aa(x1)))) |
→ |
aa(ab(ba(x1))) |
(17) |
|
cb(ba(aa(ab(x1)))) |
→ |
ca(ab(bb(x1))) |
(18) |
|
cb(ba(aa(ac(x1)))) |
→ |
ca(ab(bc(x1))) |
(19) |
|
cb(ba(aa(aa(x1)))) |
→ |
ca(ab(ba(x1))) |
(20) |
1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [bc(x1)] |
= |
1 · x1
|
| [cc(x1)] |
= |
1 + 1 · x1
|
| [ba(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [ca(x1)] |
= |
1 · x1
|
| [aa(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 · x1
|
| [ac(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
bc(cc(x1)) |
→ |
ba(ab(bc(x1))) |
(10) |
1.1.1.1.1.1 Rule Removal
Using the
matrix interpretations of dimension 2 with strict dimension 1 over the integers
| [bc(x1)] |
= |
+ · x1
|
| [ca(x1)] |
= |
+ · x1
|
| [ba(x1)] |
= |
+ · x1
|
| [ab(x1)] |
= |
+ · x1
|
| [aa(x1)] |
= |
+ · x1
|
| [bb(x1)] |
= |
+ · x1
|
| [ac(x1)] |
= |
+ · x1
|
all of the following rules can be deleted.
|
ab(ba(aa(ab(x1)))) |
→ |
aa(ab(bb(x1))) |
(15) |
|
ab(ba(aa(ac(x1)))) |
→ |
aa(ab(bc(x1))) |
(16) |
1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [bc(x1)] |
= |
1 + 1 · x1
|
| [ca(x1)] |
= |
1 · x1
|
| [ba(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [ac(x1)] |
= |
1 · x1
|
| [aa(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
bc(ca(x1)) |
→ |
ba(ab(ba(x1))) |
(11) |
1.1.1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.