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)

→

(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.

The 1^st component contains the pair

S^#(cons(x,xs))

→

S^#(xs)

(18)

1.1.1.1.1 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[S^#(x₁)]	=	3 · x₁
[cons(x₁, x₂)]	=	1 · x₁ + 3 · x₂

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair

S^#(cons(x,xs))

→

S^#(xs)

(18)

and the rules

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,s(y))	→	cons(0,int(s(0),s(y)))	(7)
int(s(x),s(y))	→	intlist(int(x,y))	(9)
intlist(nil)	→	nil	(10)
intlist(cons(x,y))	→	cons(s(x),intlist(y))	(11)
plus(plus(x,0),ext)	→	plus(x,ext)	(27)
plus(plus(x,s(y)),ext)	→	plus(s(plus(x,y)),ext)	(28)

could be deleted.

1.1.1.1.1.1 AC Dependency Pair Problem is trivial

There are no strict pairs and rules remaining, or there are no DPs remaining. Therefore, finiteness is trivially satisfied.

The 2^nd component contains the pair

plus^#(plus(x,0),ext)	→	plus^#(x,ext)	(24)
plus^#(x,s(y))	→	plus^#(x,y)	(19)
plus^#(plus(x,s(y)),ext)	→	plus^#(s(plus(x,y)),ext)	(25)
plus^#(plus(x,s(y)),ext)	→	plus^#(x,y)	(26)
plus^#(plus(x,y),z)	→	plus^#(y,z)	(16)
plus^#(x,y)	→	plus^#(y,x)	(14)
plus^#(plus(x,y),z)	→	plus^#(x,plus(y,z))	(15)

1.1.1.1.2 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[plus^#(x₁, x₂)]	=	2 · x₁ + 2 · x₂
[plus(x₁, x₂)]	=	1 · x₁ + 1 · x₂
[s(x₁)]	=	1 · x₁
[0]	=	0

together with the usable rules

plus(x,s(y))	→	s(plus(x,y))	(5)
plus(plus(x,s(y)),ext)	→	plus(s(plus(x,y)),ext)	(28)
plus(x,0)	→	x	(4)
plus(plus(x,0),ext)	→	plus(x,ext)	(27)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(13)
plus(x,y)	→	plus(y,x)	(12)

(w.r.t. the implicit argument filter of the reduction pair), the pair

plus^#(plus(x,0),ext)

→

plus^#(x,ext)

(24)

and the rules

S(cons(x,xs))	→	plus(x,S(xs))	(3)
plus(x,0)	→	x	(4)
int(0,s(y))	→	cons(0,int(s(0),s(y)))	(7)
int(s(x),s(y))	→	intlist(int(x,y))	(9)
intlist(nil)	→	nil	(10)
intlist(cons(x,y))	→	cons(s(x),intlist(y))	(11)
plus(plus(x,0),ext)	→	plus(x,ext)	(27)

could be deleted.

1.1.1.1.2.1 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[plus^#(x₁, x₂)]	=	1 · x₁ + 1 · x₂
[plus(x₁, x₂)]	=	2 + 1 · x₁ + 1 · x₂
[s(x₁)]	=	2 + 1 · x₁

together with the usable rules

plus(x,s(y))	→	s(plus(x,y))	(5)
plus(plus(x,s(y)),ext)	→	plus(s(plus(x,y)),ext)	(28)
plus(x,y)	→	plus(y,x)	(12)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(13)

(w.r.t. the implicit argument filter of the reduction pair), the pairs

plus^#(plus(x,y),z)	→	plus^#(y,z)	(16)
plus^#(plus(x,s(y)),ext)	→	plus^#(x,y)	(26)
plus^#(x,s(y))	→	plus^#(x,y)	(19)

and no rules could be deleted.

1.1.1.1.2.1.1 AC Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[plus^#(x₁, x₂)]	=	1 · x₁ + 1 · x₂
[plus(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[s(x₁)]	=	0

together with the usable rules

plus(x,s(y))	→	s(plus(x,y))	(5)
plus(plus(x,s(y)),ext)	→	plus(s(plus(x,y)),ext)	(28)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(13)
plus(x,y)	→	plus(y,x)	(12)

(w.r.t. the implicit argument filter of the reduction pair), the pair

plus^#(plus(x,s(y)),ext)

→

plus^#(s(plus(x,y)),ext)

(25)

could be deleted.

1.1.1.1.2.1.1.1 AC Dependency Pair Problem is trivial

There are no strict pairs and rules remaining, or there are no DPs remaining. Therefore, finiteness is trivially satisfied.

The 3^rd component contains the pair

int^#(s(x),s(y))	→	int^#(x,y)	(22)
int^#(0,s(y))	→	int^#(s(0),s(y))	(20)

1.1.1.1.3 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[int^#(x₁, x₂)]	=	1 · x₁ + 3 · x₂
[0]	=	0
[s(x₁)]	=	1 · x₁

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rules

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,s(y))	→	cons(0,int(s(0),s(y)))	(7)
int(s(x),s(y))	→	intlist(int(x,y))	(9)
intlist(nil)	→	nil	(10)
intlist(cons(x,y))	→	cons(s(x),intlist(y))	(11)
plus(plus(x,0),ext)	→	plus(x,ext)	(27)
plus(plus(x,s(y)),ext)	→	plus(s(plus(x,y)),ext)	(28)

could be deleted.

1.1.1.1.3.1 AC Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[int^#(x₁, x₂)]	=	3 · x₂
[0]	=	0
[s(x₁)]	=	2 + 2 · x₁

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair

int^#(s(x),s(y))

→

int^#(x,y)

(22)

could be deleted.

1.1.1.1.3.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.

The 4^th component contains the pair

intlist^#(cons(x,y))

→

intlist^#(y)

(23)

1.1.1.1.4 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[intlist^#(x₁)]	=	3 · x₁
[cons(x₁, x₂)]	=	1 · x₁ + 3 · x₂

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair

intlist^#(cons(x,y))

→

intlist^#(y)

(23)

and the rules

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,s(y))	→	cons(0,int(s(0),s(y)))	(7)
int(s(x),s(y))	→	intlist(int(x,y))	(9)
intlist(nil)	→	nil	(10)
intlist(cons(x,y))	→	cons(s(x),intlist(y))	(11)
plus(plus(x,0),ext)	→	plus(x,ext)	(27)
plus(plus(x,s(y)),ext)	→	plus(s(plus(x,y)),ext)	(28)

could be deleted.

1.1.1.1.4.1 AC Dependency Pair Problem is trivial

There are no strict pairs and rules remaining, or there are no DPs remaining. Therefore, finiteness is trivially satisfied.