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 ttt2 @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
b(a(x1)) |
→ |
C(x1) |
(25) |
|
c(b(x1)) |
→ |
A(x1) |
(26) |
|
a(c(x1)) |
→ |
B(x1) |
(27) |
|
C(A(x1)) |
→ |
b(x1) |
(28) |
|
B(C(x1)) |
→ |
a(x1) |
(29) |
|
A(B(x1)) |
→ |
c(x1) |
(30) |
|
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(a(B(x1)))) |
→ |
A(A(A(c(x1)))) |
(31) |
|
b(A(A(A(x1)))) |
→ |
C(a(a(a(x1)))) |
(32) |
|
b(b(b(C(x1)))) |
→ |
B(B(B(a(x1)))) |
(33) |
|
c(B(B(B(x1)))) |
→ |
A(b(b(b(x1)))) |
(34) |
|
c(c(c(A(x1)))) |
→ |
C(C(C(b(x1)))) |
(35) |
|
a(C(C(C(x1)))) |
→ |
B(c(c(c(x1)))) |
(36) |
|
A(a(x1)) |
→ |
x1 |
(20) |
|
a(A(x1)) |
→ |
x1 |
(19) |
|
B(b(x1)) |
→ |
x1 |
(22) |
|
b(B(x1)) |
→ |
x1 |
(21) |
|
C(c(x1)) |
→ |
x1 |
(24) |
|
c(C(x1)) |
→ |
x1 |
(23) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(B) |
= |
0 |
|
weight(B) |
= |
1 |
|
|
|
| prec(A) |
= |
2 |
|
weight(A) |
= |
1 |
|
|
|
| prec(c) |
= |
4 |
|
weight(c) |
= |
1 |
|
|
|
| prec(C) |
= |
1 |
|
weight(C) |
= |
1 |
|
|
|
| prec(a) |
= |
5 |
|
weight(a) |
= |
1 |
|
|
|
| prec(b) |
= |
3 |
|
weight(b) |
= |
1 |
|
|
|
all of the following rules can be deleted.
|
b(a(x1)) |
→ |
C(x1) |
(25) |
|
c(b(x1)) |
→ |
A(x1) |
(26) |
|
a(c(x1)) |
→ |
B(x1) |
(27) |
|
C(A(x1)) |
→ |
b(x1) |
(28) |
|
B(C(x1)) |
→ |
a(x1) |
(29) |
|
A(B(x1)) |
→ |
c(x1) |
(30) |
|
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(a(B(x1)))) |
→ |
A(A(A(c(x1)))) |
(31) |
|
b(A(A(A(x1)))) |
→ |
C(a(a(a(x1)))) |
(32) |
|
b(b(b(C(x1)))) |
→ |
B(B(B(a(x1)))) |
(33) |
|
c(B(B(B(x1)))) |
→ |
A(b(b(b(x1)))) |
(34) |
|
c(c(c(A(x1)))) |
→ |
C(C(C(b(x1)))) |
(35) |
|
a(C(C(C(x1)))) |
→ |
B(c(c(c(x1)))) |
(36) |
|
A(a(x1)) |
→ |
x1 |
(20) |
|
a(A(x1)) |
→ |
x1 |
(19) |
|
B(b(x1)) |
→ |
x1 |
(22) |
|
b(B(x1)) |
→ |
x1 |
(21) |
|
C(c(x1)) |
→ |
x1 |
(24) |
|
c(C(x1)) |
→ |
x1 |
(23) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.