Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/MYNAT_nosorts_noand_GM)
The rewrite relation of the following TRS is considered.
a__U11(tt,M,N) |
→ |
a__U12(tt,M,N) |
(1) |
a__U12(tt,M,N) |
→ |
s(a__plus(mark(N),mark(M))) |
(2) |
a__U21(tt,M,N) |
→ |
a__U22(tt,M,N) |
(3) |
a__U22(tt,M,N) |
→ |
a__plus(a__x(mark(N),mark(M)),mark(N)) |
(4) |
a__plus(N,0) |
→ |
mark(N) |
(5) |
a__plus(N,s(M)) |
→ |
a__U11(tt,M,N) |
(6) |
a__x(N,0) |
→ |
0 |
(7) |
a__x(N,s(M)) |
→ |
a__U21(tt,M,N) |
(8) |
mark(U11(X1,X2,X3)) |
→ |
a__U11(mark(X1),X2,X3) |
(9) |
mark(U12(X1,X2,X3)) |
→ |
a__U12(mark(X1),X2,X3) |
(10) |
mark(plus(X1,X2)) |
→ |
a__plus(mark(X1),mark(X2)) |
(11) |
mark(U21(X1,X2,X3)) |
→ |
a__U21(mark(X1),X2,X3) |
(12) |
mark(U22(X1,X2,X3)) |
→ |
a__U22(mark(X1),X2,X3) |
(13) |
mark(x(X1,X2)) |
→ |
a__x(mark(X1),mark(X2)) |
(14) |
mark(tt) |
→ |
tt |
(15) |
mark(s(X)) |
→ |
s(mark(X)) |
(16) |
mark(0) |
→ |
0 |
(17) |
a__U11(X1,X2,X3) |
→ |
U11(X1,X2,X3) |
(18) |
a__U12(X1,X2,X3) |
→ |
U12(X1,X2,X3) |
(19) |
a__plus(X1,X2) |
→ |
plus(X1,X2) |
(20) |
a__U21(X1,X2,X3) |
→ |
U21(X1,X2,X3) |
(21) |
a__U22(X1,X2,X3) |
→ |
U22(X1,X2,X3) |
(22) |
a__x(X1,X2) |
→ |
x(X1,X2) |
(23) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
prec(a__U11) |
= |
1 |
|
stat(a__U11) |
= |
mul
|
prec(tt) |
= |
0 |
|
stat(tt) |
= |
mul
|
prec(a__U12) |
= |
1 |
|
stat(a__U12) |
= |
mul
|
prec(s) |
= |
0 |
|
stat(s) |
= |
mul
|
prec(a__plus) |
= |
1 |
|
stat(a__plus) |
= |
mul
|
prec(a__U21) |
= |
2 |
|
stat(a__U21) |
= |
lex
|
prec(a__U22) |
= |
2 |
|
stat(a__U22) |
= |
lex
|
prec(a__x) |
= |
2 |
|
stat(a__x) |
= |
lex
|
prec(0) |
= |
3 |
|
stat(0) |
= |
mul
|
prec(U11) |
= |
1 |
|
stat(U11) |
= |
mul
|
prec(U12) |
= |
1 |
|
stat(U12) |
= |
mul
|
prec(plus) |
= |
1 |
|
stat(plus) |
= |
mul
|
prec(U21) |
= |
2 |
|
stat(U21) |
= |
lex
|
prec(U22) |
= |
2 |
|
stat(U22) |
= |
lex
|
prec(x) |
= |
2 |
|
stat(x) |
= |
lex
|
π(a__U11) |
= |
[1,2,3] |
π(tt) |
= |
[] |
π(a__U12) |
= |
[1,2,3] |
π(s) |
= |
[1] |
π(a__plus) |
= |
[1,2] |
π(mark) |
= |
1 |
π(a__U21) |
= |
[3,2,1] |
π(a__U22) |
= |
[3,2,1] |
π(a__x) |
= |
[1,2] |
π(0) |
= |
[] |
π(U11) |
= |
[1,2,3] |
π(U12) |
= |
[1,2,3] |
π(plus) |
= |
[1,2] |
π(U21) |
= |
[3,2,1] |
π(U22) |
= |
[3,2,1] |
π(x) |
= |
[1,2] |
all of the following rules can be deleted.
a__U12(tt,M,N) |
→ |
s(a__plus(mark(N),mark(M))) |
(2) |
a__U22(tt,M,N) |
→ |
a__plus(a__x(mark(N),mark(M)),mark(N)) |
(4) |
a__plus(N,0) |
→ |
mark(N) |
(5) |
a__plus(N,s(M)) |
→ |
a__U11(tt,M,N) |
(6) |
a__x(N,0) |
→ |
0 |
(7) |
a__x(N,s(M)) |
→ |
a__U21(tt,M,N) |
(8) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(tt) |
= |
1 |
|
weight(tt) |
= |
1 |
|
|
|
prec(0) |
= |
14 |
|
weight(0) |
= |
1 |
|
|
|
prec(mark) |
= |
15 |
|
weight(mark) |
= |
0 |
|
|
|
prec(s) |
= |
7 |
|
weight(s) |
= |
1 |
|
|
|
prec(a__U11) |
= |
11 |
|
weight(a__U11) |
= |
0 |
|
|
|
prec(a__U12) |
= |
10 |
|
weight(a__U12) |
= |
0 |
|
|
|
prec(a__U21) |
= |
3 |
|
weight(a__U21) |
= |
1 |
|
|
|
prec(a__U22) |
= |
13 |
|
weight(a__U22) |
= |
0 |
|
|
|
prec(U11) |
= |
0 |
|
weight(U11) |
= |
0 |
|
|
|
prec(U12) |
= |
9 |
|
weight(U12) |
= |
0 |
|
|
|
prec(plus) |
= |
4 |
|
weight(plus) |
= |
0 |
|
|
|
prec(a__plus) |
= |
5 |
|
weight(a__plus) |
= |
0 |
|
|
|
prec(U21) |
= |
2 |
|
weight(U21) |
= |
1 |
|
|
|
prec(U22) |
= |
12 |
|
weight(U22) |
= |
0 |
|
|
|
prec(x) |
= |
6 |
|
weight(x) |
= |
0 |
|
|
|
prec(a__x) |
= |
8 |
|
weight(a__x) |
= |
0 |
|
|
|
all of the following rules can be deleted.
a__U11(tt,M,N) |
→ |
a__U12(tt,M,N) |
(1) |
a__U21(tt,M,N) |
→ |
a__U22(tt,M,N) |
(3) |
mark(U11(X1,X2,X3)) |
→ |
a__U11(mark(X1),X2,X3) |
(9) |
mark(U12(X1,X2,X3)) |
→ |
a__U12(mark(X1),X2,X3) |
(10) |
mark(plus(X1,X2)) |
→ |
a__plus(mark(X1),mark(X2)) |
(11) |
mark(U21(X1,X2,X3)) |
→ |
a__U21(mark(X1),X2,X3) |
(12) |
mark(U22(X1,X2,X3)) |
→ |
a__U22(mark(X1),X2,X3) |
(13) |
mark(x(X1,X2)) |
→ |
a__x(mark(X1),mark(X2)) |
(14) |
mark(tt) |
→ |
tt |
(15) |
mark(s(X)) |
→ |
s(mark(X)) |
(16) |
mark(0) |
→ |
0 |
(17) |
a__U11(X1,X2,X3) |
→ |
U11(X1,X2,X3) |
(18) |
a__U12(X1,X2,X3) |
→ |
U12(X1,X2,X3) |
(19) |
a__plus(X1,X2) |
→ |
plus(X1,X2) |
(20) |
a__U21(X1,X2,X3) |
→ |
U21(X1,X2,X3) |
(21) |
a__U22(X1,X2,X3) |
→ |
U22(X1,X2,X3) |
(22) |
a__x(X1,X2) |
→ |
x(X1,X2) |
(23) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.