Certification Problem

Input (TPDB TRS_Standard/TCT_12/recursion-5)

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)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the Weighted Path Order with the following precedence and status
prec(g_5) = 13 status(g_5) = [2, 1] list-extension(g_5) = Lex
prec(f_5) = 0 status(f_5) = [1] list-extension(f_5) = Lex
prec(g_4) = 9 status(g_4) = [1, 2] list-extension(g_4) = Lex
prec(f_4) = 12 status(f_4) = [1] list-extension(f_4) = Lex
prec(g_3) = 11 status(g_3) = [1, 2] list-extension(g_3) = Lex
prec(f_3) = 0 status(f_3) = [1] list-extension(f_3) = Lex
prec(g_2) = 8 status(g_2) = [2, 1] list-extension(g_2) = Lex
prec(f_2) = 9 status(f_2) = [1] list-extension(f_2) = Lex
prec(b) = 0 status(b) = [2, 1] list-extension(b) = Lex
prec(s) = 0 status(s) = [1] list-extension(s) = Lex
prec(g_1) = 2 status(g_1) = [2, 1] list-extension(g_1) = Lex
prec(f_1) = 4 status(f_1) = [1] list-extension(f_1) = Lex
prec(a) = 0 status(a) = [] list-extension(a) = Lex
prec(f_0) = 1 status(f_0) = [1] list-extension(f_0) = Lex
and the following Max-polynomial interpretation
[g_5(x1, x2)] = max(2, 2 + 1 · x1, 1 + 1 · x2)
[f_5(x1)] = 7 + 1 · x1
[g_4(x1, x2)] = max(1, 1 + 1 · x1, 1 + 1 · x2)
[f_4(x1)] = max(0, 1 + 1 · x1)
[g_3(x1, x2)] = max(0, 0 + 1 · x1, 0 + 1 · x2)
[f_3(x1)] = 1 + 1 · x1
[g_2(x1, x2)] = max(0, 0 + 1 · x1, 0 + 1 · x2)
[f_2(x1)] = max(0, 0 + 1 · x1)
[b(x1, x2)] = max(0, 0 + 1 · x1, 0 + 1 · x2)
[s(x1)] = 0 + 1 · x1
[g_1(x1, x2)] = max(0, 0 + 1 · x1, 0 + 1 · x2)
[f_1(x1)] = max(0, 0 + 1 · x1)
[a] = max(0)
[f_0(x1)] = 0 + 1 · x1
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)

1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.