Certification Problem

Input (TPDB TRS_Equational/AProVE_AC_04/AC07)

The rewrite relation of the following equational TRS is considered.

sum(x,y) S(int(x,y)) (1)
S(nil) 0 (2)
S(cons(x,xs)) plus(x,S(xs)) (3)
plus(x,0) x (4)
plus(x,s(y)) s(plus(x,y)) (5)
int(0,0) cons(0,nil) (6)
int(0,s(y)) cons(0,int(s(0),s(y))) (7)
int(s(x),0) nil (8)
int(s(x),s(y)) intlist(int(x,y)) (9)
intlist(nil) nil (10)
intlist(cons(x,y)) cons(s(x),intlist(y)) (11)

Associative symbols: plus

Commutative symbols: plus

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 AC Rule Removal

Using the non-linear polynomial interpretation over the naturals
[plus(x1, x2)] = 1 · x2 + 1 · x1
[sum(x1, x2)] = 2 + 1 · x2 + 1 · x1 + 1 · x1 · x2
[S(x1)] = 1 · x1
[int(x1, x2)] = 1 + 1 · x2 + 1 · x1
[nil] = 0
[0] = 0
[cons(x1, x2)] = 1 · x2 + 2 · x1
[s(x1)] = 1 · x1
[intlist(x1)] = 1 · x1
the rules
sum(x,y) S(int(x,y)) (1)
int(0,0) cons(0,nil) (6)
int(s(x),0) nil (8)
can be deleted.

1.1 AC Rule Removal

Using the linear polynomial interpretation over the naturals
[plus(x1, x2)] = 1 · x2 + 1 · x1
[S(x1)] = 2 · x1
[nil] = 1
[0] = 0
[cons(x1, x2)] = 1 · x2 + 2 · x1
[s(x1)] = 1 · x1
[int(x1, x2)] = 1 · x2 + 2 · x1
[intlist(x1)] = 1 · x1
the rule
S(nil) 0 (2)
can be deleted.

1.1.1 AC Dependency Pair Transformation

The following set of (strict) dependency pairs is constructed for the TRS.

S#(cons(x,xs)) plus#(x,S(xs)) (17)
S#(cons(x,xs)) S#(xs) (18)
plus#(x,s(y)) plus#(x,y) (19)
int#(0,s(y)) int#(s(0),s(y)) (20)
int#(s(x),s(y)) intlist#(int(x,y)) (21)
int#(s(x),s(y)) int#(x,y) (22)
intlist#(cons(x,y)) intlist#(y) (23)
plus#(plus(x,0),ext) plus#(x,ext) (24)
plus#(plus(x,s(y)),ext) plus#(s(plus(x,y)),ext) (25)
plus#(plus(x,s(y)),ext) plus#(x,y) (26)
The extended rules of the TRS
plus(plus(x,0),ext) plus(x,ext) (27)
plus(plus(x,s(y)),ext) plus(s(plus(x,y)),ext) (28)
give rise to even more dependency pairs (by sharping the root symbols of each rule). Finiteness for all DPs in combination with the equational DPs is proven as follows.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.