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 NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
f_3#(x) g_3#(x,x) (12)
g_5#(s(x),y) g_5#(x,y) (13)
g_1#(s(x),y) g_1#(x,y) (14)
f_5#(x) g_5#(x,x) (15)
g_2#(s(x),y) g_2#(x,y) (16)
g_1#(s(x),y) f_0#(y) (17)
f_4#(x) g_4#(x,x) (18)
g_3#(s(x),y) f_2#(y) (19)
g_2#(s(x),y) f_1#(y) (20)
g_5#(s(x),y) f_4#(y) (21)
g_4#(s(x),y) f_3#(y) (22)
f_1#(x) g_1#(x,x) (23)
g_3#(s(x),y) g_3#(x,y) (24)
f_2#(x) g_2#(x,x) (25)
g_4#(s(x),y) g_4#(x,y) (26)

1.1 Dependency Graph Processor

The dependency pairs are split into 5 components.