Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z069)
The rewrite relation of the following TRS is considered.
|
a(b(x1)) |
→ |
C(x1) |
(1) |
|
b(c(x1)) |
→ |
A(x1) |
(2) |
|
c(a(x1)) |
→ |
B(x1) |
(3) |
|
A(C(x1)) |
→ |
b(x1) |
(4) |
|
C(B(x1)) |
→ |
a(x1) |
(5) |
|
B(A(x1)) |
→ |
c(x1) |
(6) |
|
a(a(a(a(x1)))) |
→ |
A(A(A(x1))) |
(7) |
|
A(A(A(A(x1)))) |
→ |
a(a(a(x1))) |
(8) |
|
b(b(b(b(x1)))) |
→ |
B(B(B(x1))) |
(9) |
|
B(B(B(B(x1)))) |
→ |
b(b(b(x1))) |
(10) |
|
c(c(c(c(x1)))) |
→ |
C(C(C(x1))) |
(11) |
|
C(C(C(C(x1)))) |
→ |
c(c(c(x1))) |
(12) |
|
B(a(a(a(x1)))) |
→ |
c(A(A(A(x1)))) |
(13) |
|
A(A(A(b(x1)))) |
→ |
a(a(a(C(x1)))) |
(14) |
|
C(b(b(b(x1)))) |
→ |
a(B(B(B(x1)))) |
(15) |
|
B(B(B(c(x1)))) |
→ |
b(b(b(A(x1)))) |
(16) |
|
A(c(c(c(x1)))) |
→ |
b(C(C(C(x1)))) |
(17) |
|
C(C(C(a(x1)))) |
→ |
c(c(c(B(x1)))) |
(18) |
|
a(A(x1)) |
→ |
x1 |
(19) |
|
A(a(x1)) |
→ |
x1 |
(20) |
|
b(B(x1)) |
→ |
x1 |
(21) |
|
B(b(x1)) |
→ |
x1 |
(22) |
|
c(C(x1)) |
→ |
x1 |
(23) |
|
C(c(x1)) |
→ |
x1 |
(24) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [a(x1)] |
= |
1 · x1 + 1 |
| [b(x1)] |
= |
1 · x1 + 1 |
| [C(x1)] |
= |
1 · x1 + 1 |
| [c(x1)] |
= |
1 · x1 + 1 |
| [A(x1)] |
= |
1 · x1 + 1 |
| [B(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
|
a(b(x1)) |
→ |
C(x1) |
(1) |
|
b(c(x1)) |
→ |
A(x1) |
(2) |
|
c(a(x1)) |
→ |
B(x1) |
(3) |
|
A(C(x1)) |
→ |
b(x1) |
(4) |
|
C(B(x1)) |
→ |
a(x1) |
(5) |
|
B(A(x1)) |
→ |
c(x1) |
(6) |
|
a(a(a(a(x1)))) |
→ |
A(A(A(x1))) |
(7) |
|
A(A(A(A(x1)))) |
→ |
a(a(a(x1))) |
(8) |
|
b(b(b(b(x1)))) |
→ |
B(B(B(x1))) |
(9) |
|
B(B(B(B(x1)))) |
→ |
b(b(b(x1))) |
(10) |
|
c(c(c(c(x1)))) |
→ |
C(C(C(x1))) |
(11) |
|
C(C(C(C(x1)))) |
→ |
c(c(c(x1))) |
(12) |
|
a(A(x1)) |
→ |
x1 |
(19) |
|
A(a(x1)) |
→ |
x1 |
(20) |
|
b(B(x1)) |
→ |
x1 |
(21) |
|
B(b(x1)) |
→ |
x1 |
(22) |
|
c(C(x1)) |
→ |
x1 |
(23) |
|
C(c(x1)) |
→ |
x1 |
(24) |
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 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
B#(a(a(a(x1)))) |
→ |
A#(A(A(x1))) |
(25) |
|
B#(a(a(a(x1)))) |
→ |
A#(A(x1)) |
(26) |
|
B#(a(a(a(x1)))) |
→ |
A#(x1) |
(27) |
|
A#(A(A(b(x1)))) |
→ |
C#(x1) |
(28) |
|
C#(b(b(b(x1)))) |
→ |
B#(B(B(x1))) |
(29) |
|
C#(b(b(b(x1)))) |
→ |
B#(B(x1)) |
(30) |
|
C#(b(b(b(x1)))) |
→ |
B#(x1) |
(31) |
|
B#(B(B(c(x1)))) |
→ |
A#(x1) |
(32) |
|
A#(c(c(c(x1)))) |
→ |
C#(C(C(x1))) |
(33) |
|
A#(c(c(c(x1)))) |
→ |
C#(C(x1)) |
(34) |
|
A#(c(c(c(x1)))) |
→ |
C#(x1) |
(35) |
|
C#(C(C(a(x1)))) |
→ |
B#(x1) |
(36) |
1.1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [B#(x1)] |
= |
1 + 1 · x1
|
| [a(x1)] |
= |
1 + 1 · x1
|
| [A#(x1)] |
= |
1 + 1 · x1
|
| [A(x1)] |
= |
1 + 1 · x1
|
| [b(x1)] |
= |
1 + 1 · x1
|
| [C#(x1)] |
= |
1 + 1 · x1
|
| [B(x1)] |
= |
1 + 1 · x1
|
| [c(x1)] |
= |
1 + 1 · x1
|
| [C(x1)] |
= |
1 + 1 · x1
|
the
pairs
|
B#(a(a(a(x1)))) |
→ |
A#(A(A(x1))) |
(25) |
|
B#(a(a(a(x1)))) |
→ |
A#(A(x1)) |
(26) |
|
B#(a(a(a(x1)))) |
→ |
A#(x1) |
(27) |
|
A#(A(A(b(x1)))) |
→ |
C#(x1) |
(28) |
|
C#(b(b(b(x1)))) |
→ |
B#(B(B(x1))) |
(29) |
|
C#(b(b(b(x1)))) |
→ |
B#(B(x1)) |
(30) |
|
C#(b(b(b(x1)))) |
→ |
B#(x1) |
(31) |
|
B#(B(B(c(x1)))) |
→ |
A#(x1) |
(32) |
|
A#(c(c(c(x1)))) |
→ |
C#(C(C(x1))) |
(33) |
|
A#(c(c(c(x1)))) |
→ |
C#(C(x1)) |
(34) |
|
A#(c(c(c(x1)))) |
→ |
C#(x1) |
(35) |
|
C#(C(C(a(x1)))) |
→ |
B#(x1) |
(36) |
could be deleted.
1.1.1.1.1 P is empty
There are no pairs anymore.