Certification Problem
Input (TPDB SRS_Relative/Mixed_relative_SRS/zr11)
The relative rewrite relation R/S is considered where R is the following TRS
|
b(c(a(x1))) |
→ |
a(b(a(b(x1)))) |
(1) |
|
a(a(x1)) |
→ |
a(c(b(a(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
|
a(c(b(x1))) |
→ |
b(a(b(a(x1)))) |
(4) |
|
a(a(x1)) |
→ |
a(b(c(a(x1)))) |
(5) |
|
b(x1) |
→ |
b(c(x1)) |
(6) |
1.1 Closure Under Flat Contexts
Using the flat contexts
{a(☐), c(☐), b(☐)}
We obtain the transformed TRS
|
a(a(x1)) |
→ |
a(b(c(a(x1)))) |
(5) |
|
a(a(c(b(x1)))) |
→ |
a(b(a(b(a(x1))))) |
(7) |
|
c(a(c(b(x1)))) |
→ |
c(b(a(b(a(x1))))) |
(8) |
|
b(a(c(b(x1)))) |
→ |
b(b(a(b(a(x1))))) |
(9) |
|
b(x1) |
→ |
b(c(x1)) |
(6) |
1.1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
aa(aa(x1)) |
→ |
ab(bc(ca(aa(x1)))) |
(10) |
|
aa(ab(x1)) |
→ |
ab(bc(ca(ab(x1)))) |
(11) |
|
aa(ac(x1)) |
→ |
ab(bc(ca(ac(x1)))) |
(12) |
|
aa(ac(cb(ba(x1)))) |
→ |
ab(ba(ab(ba(aa(x1))))) |
(13) |
|
aa(ac(cb(bb(x1)))) |
→ |
ab(ba(ab(ba(ab(x1))))) |
(14) |
|
aa(ac(cb(bc(x1)))) |
→ |
ab(ba(ab(ba(ac(x1))))) |
(15) |
|
ca(ac(cb(ba(x1)))) |
→ |
cb(ba(ab(ba(aa(x1))))) |
(16) |
|
ca(ac(cb(bb(x1)))) |
→ |
cb(ba(ab(ba(ab(x1))))) |
(17) |
|
ca(ac(cb(bc(x1)))) |
→ |
cb(ba(ab(ba(ac(x1))))) |
(18) |
|
ba(ac(cb(ba(x1)))) |
→ |
bb(ba(ab(ba(aa(x1))))) |
(19) |
|
ba(ac(cb(bb(x1)))) |
→ |
bb(ba(ab(ba(ab(x1))))) |
(20) |
|
ba(ac(cb(bc(x1)))) |
→ |
bb(ba(ab(ba(ac(x1))))) |
(21) |
|
ba(x1) |
→ |
bc(ca(x1)) |
(22) |
|
bb(x1) |
→ |
bc(cb(x1)) |
(23) |
|
bc(x1) |
→ |
bc(cc(x1)) |
(24) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [aa(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [bc(x1)] |
= |
1 · x1
|
| [ca(x1)] |
= |
1 · x1
|
| [ac(x1)] |
= |
1 + 1 · x1
|
| [cb(x1)] |
= |
1 · x1
|
| [ba(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 · x1
|
| [cc(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
aa(ac(cb(ba(x1)))) |
→ |
ab(ba(ab(ba(aa(x1))))) |
(13) |
|
aa(ac(cb(bb(x1)))) |
→ |
ab(ba(ab(ba(ab(x1))))) |
(14) |
|
ca(ac(cb(ba(x1)))) |
→ |
cb(ba(ab(ba(aa(x1))))) |
(16) |
|
ca(ac(cb(bb(x1)))) |
→ |
cb(ba(ab(ba(ab(x1))))) |
(17) |
|
ba(ac(cb(ba(x1)))) |
→ |
bb(ba(ab(ba(aa(x1))))) |
(19) |
|
ba(ac(cb(bb(x1)))) |
→ |
bb(ba(ab(ba(ab(x1))))) |
(20) |
1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [aa(x1)] |
= |
1 + 1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [bc(x1)] |
= |
1 · x1
|
| [ca(x1)] |
= |
1 · x1
|
| [ac(x1)] |
= |
1 · x1
|
| [cb(x1)] |
= |
1 · x1
|
| [ba(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 · x1
|
| [cc(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
aa(aa(x1)) |
→ |
ab(bc(ca(aa(x1)))) |
(10) |
|
aa(ab(x1)) |
→ |
ab(bc(ca(ab(x1)))) |
(11) |
|
aa(ac(x1)) |
→ |
ab(bc(ca(ac(x1)))) |
(12) |
|
aa(ac(cb(bc(x1)))) |
→ |
ab(ba(ab(ba(ac(x1))))) |
(15) |
1.1.1.1.1.1 Rule Removal
Using the
matrix interpretations of dimension 2 with strict dimension 1 over the integers
| [ca(x1)] |
= |
+ · x1
|
| [ac(x1)] |
= |
+ · x1
|
| [cb(x1)] |
= |
+ · x1
|
| [bc(x1)] |
= |
+ · x1
|
| [ba(x1)] |
= |
+ · x1
|
| [ab(x1)] |
= |
+ · x1
|
| [bb(x1)] |
= |
+ · x1
|
| [cc(x1)] |
= |
+ · x1
|
all of the following rules can be deleted.
|
ca(ac(cb(bc(x1)))) |
→ |
cb(ba(ab(ba(ac(x1))))) |
(18) |
|
bb(x1) |
→ |
bc(cb(x1)) |
(23) |
1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [ba(x1)] |
= |
1 + 1 · x1
|
| [ac(x1)] |
= |
1 · x1
|
| [cb(x1)] |
= |
1 + 1 · x1
|
| [bc(x1)] |
= |
1 + 1 · x1
|
| [bb(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [ca(x1)] |
= |
1 · x1
|
| [cc(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
ba(ac(cb(bc(x1)))) |
→ |
bb(ba(ab(ba(ac(x1))))) |
(21) |
1.1.1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.