Certification Problem
Input (TPDB TRS_Standard/TCT_12/recursion-10)
The rewrite relation of the following TRS is considered.
|
f_0(x) |
→ |
a |
(1) |
|
f_1(x) |
→ |
g_1(x,x) |
(2) |
|
g_1(s(x),y) |
→ |
b(f_0(y),g_1(x,y)) |
(3) |
|
f_2(x) |
→ |
g_2(x,x) |
(4) |
|
g_2(s(x),y) |
→ |
b(f_1(y),g_2(x,y)) |
(5) |
|
f_3(x) |
→ |
g_3(x,x) |
(6) |
|
g_3(s(x),y) |
→ |
b(f_2(y),g_3(x,y)) |
(7) |
|
f_4(x) |
→ |
g_4(x,x) |
(8) |
|
g_4(s(x),y) |
→ |
b(f_3(y),g_4(x,y)) |
(9) |
|
f_5(x) |
→ |
g_5(x,x) |
(10) |
|
g_5(s(x),y) |
→ |
b(f_4(y),g_5(x,y)) |
(11) |
|
f_6(x) |
→ |
g_6(x,x) |
(12) |
|
g_6(s(x),y) |
→ |
b(f_5(y),g_6(x,y)) |
(13) |
|
f_7(x) |
→ |
g_7(x,x) |
(14) |
|
g_7(s(x),y) |
→ |
b(f_6(y),g_7(x,y)) |
(15) |
|
f_8(x) |
→ |
g_8(x,x) |
(16) |
|
g_8(s(x),y) |
→ |
b(f_7(y),g_8(x,y)) |
(17) |
|
f_9(x) |
→ |
g_9(x,x) |
(18) |
|
g_9(s(x),y) |
→ |
b(f_8(y),g_9(x,y)) |
(19) |
|
f_10(x) |
→ |
g_10(x,x) |
(20) |
|
g_10(s(x),y) |
→ |
b(f_9(y),g_10(x,y)) |
(21) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(f_0) |
= |
0 |
|
stat(f_0) |
= |
mul
|
| prec(a) |
= |
0 |
|
stat(a) |
= |
mul
|
| prec(f_1) |
= |
3 |
|
stat(f_1) |
= |
mul
|
| prec(g_1) |
= |
2 |
|
stat(g_1) |
= |
mul
|
| prec(s) |
= |
6 |
|
stat(s) |
= |
mul
|
| prec(b) |
= |
1 |
|
stat(b) |
= |
mul
|
| prec(f_2) |
= |
4 |
|
stat(f_2) |
= |
mul
|
| prec(g_2) |
= |
3 |
|
stat(g_2) |
= |
mul
|
| prec(f_3) |
= |
5 |
|
stat(f_3) |
= |
mul
|
| prec(g_3) |
= |
4 |
|
stat(g_3) |
= |
mul
|
| prec(f_4) |
= |
7 |
|
stat(f_4) |
= |
mul
|
| prec(g_4) |
= |
5 |
|
stat(g_4) |
= |
mul
|
| prec(f_5) |
= |
8 |
|
stat(f_5) |
= |
mul
|
| prec(g_5) |
= |
7 |
|
stat(g_5) |
= |
mul
|
| prec(f_6) |
= |
9 |
|
stat(f_6) |
= |
mul
|
| prec(g_6) |
= |
8 |
|
stat(g_6) |
= |
mul
|
| prec(f_7) |
= |
10 |
|
stat(f_7) |
= |
mul
|
| prec(g_7) |
= |
9 |
|
stat(g_7) |
= |
mul
|
| prec(f_8) |
= |
11 |
|
stat(f_8) |
= |
mul
|
| prec(g_8) |
= |
10 |
|
stat(g_8) |
= |
mul
|
| prec(f_9) |
= |
13 |
|
stat(f_9) |
= |
mul
|
| prec(g_9) |
= |
12 |
|
stat(g_9) |
= |
mul
|
| prec(f_10) |
= |
14 |
|
stat(f_10) |
= |
mul
|
| prec(g_10) |
= |
13 |
|
stat(g_10) |
= |
mul
|
| π(f_0) |
= |
[1] |
| π(a) |
= |
[] |
| π(f_1) |
= |
[1] |
| π(g_1) |
= |
[1,2] |
| π(s) |
= |
[1] |
| π(b) |
= |
[1,2] |
| π(f_2) |
= |
[1] |
| π(g_2) |
= |
[1,2] |
| π(f_3) |
= |
[1] |
| π(g_3) |
= |
[1,2] |
| π(f_4) |
= |
[1] |
| π(g_4) |
= |
[1,2] |
| π(f_5) |
= |
[1] |
| π(g_5) |
= |
[1,2] |
| π(f_6) |
= |
[1] |
| π(g_6) |
= |
[1,2] |
| π(f_7) |
= |
[1] |
| π(g_7) |
= |
[1,2] |
| π(f_8) |
= |
[1] |
| π(g_8) |
= |
[1,2] |
| π(f_9) |
= |
[1] |
| π(g_9) |
= |
[1,2] |
| π(f_10) |
= |
[1] |
| π(g_10) |
= |
[1,2] |
all of the following rules can be deleted.
|
f_0(x) |
→ |
a |
(1) |
|
f_1(x) |
→ |
g_1(x,x) |
(2) |
|
g_1(s(x),y) |
→ |
b(f_0(y),g_1(x,y)) |
(3) |
|
f_2(x) |
→ |
g_2(x,x) |
(4) |
|
g_2(s(x),y) |
→ |
b(f_1(y),g_2(x,y)) |
(5) |
|
f_3(x) |
→ |
g_3(x,x) |
(6) |
|
g_3(s(x),y) |
→ |
b(f_2(y),g_3(x,y)) |
(7) |
|
f_4(x) |
→ |
g_4(x,x) |
(8) |
|
g_4(s(x),y) |
→ |
b(f_3(y),g_4(x,y)) |
(9) |
|
f_5(x) |
→ |
g_5(x,x) |
(10) |
|
g_5(s(x),y) |
→ |
b(f_4(y),g_5(x,y)) |
(11) |
|
f_6(x) |
→ |
g_6(x,x) |
(12) |
|
g_6(s(x),y) |
→ |
b(f_5(y),g_6(x,y)) |
(13) |
|
f_7(x) |
→ |
g_7(x,x) |
(14) |
|
g_7(s(x),y) |
→ |
b(f_6(y),g_7(x,y)) |
(15) |
|
f_8(x) |
→ |
g_8(x,x) |
(16) |
|
g_8(s(x),y) |
→ |
b(f_7(y),g_8(x,y)) |
(17) |
|
f_9(x) |
→ |
g_9(x,x) |
(18) |
|
g_9(s(x),y) |
→ |
b(f_8(y),g_9(x,y)) |
(19) |
|
f_10(x) |
→ |
g_10(x,x) |
(20) |
|
g_10(s(x),y) |
→ |
b(f_9(y),g_10(x,y)) |
(21) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.