Certification Problem

Input (TPDB TRS_Equational/AProVE_AC_04/AC54)

The rewrite relation of the following equational TRS is considered.

minus(x,0)	→	x	(1)
minus(s(x),s(y))	→	minus(x,y)	(2)
minus(minus(x,y),z)	→	minus(x,plus(y,z))	(3)
quot(0,s(y))	→	0	(4)
quot(s(x),s(y))	→	s(quot(minus(x,y),s(y)))	(5)
plus(0,y)	→	y	(6)
plus(s(x),y)	→	s(plus(x,y))	(7)
app(nil,k)	→	k	(8)
app(l,nil)	→	l	(9)
app(cons(x,l),k)	→	cons(x,app(l,k))	(10)
sum(cons(x,nil))	→	cons(x,nil)	(11)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(12)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(13)

Associative symbols: plus

Commutative symbols: plus

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 AC Dependency Pair Transformation

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

minus^#(s(x),s(y))	→	minus^#(x,y)	(19)
minus^#(minus(x,y),z)	→	minus^#(x,plus(y,z))	(20)
minus^#(minus(x,y),z)	→	plus^#(y,z)	(21)
quot^#(s(x),s(y))	→	quot^#(minus(x,y),s(y))	(22)
quot^#(s(x),s(y))	→	minus^#(x,y)	(23)
plus^#(s(x),y)	→	plus^#(x,y)	(24)
app^#(cons(x,l),k)	→	app^#(l,k)	(25)
sum^#(cons(x,cons(y,l)))	→	sum^#(cons(plus(x,y),l))	(26)
sum^#(cons(x,cons(y,l)))	→	plus^#(x,y)	(27)
sum^#(app(l,cons(x,cons(y,k))))	→	sum^#(app(l,sum(cons(x,cons(y,k)))))	(28)
sum^#(app(l,cons(x,cons(y,k))))	→	app^#(l,sum(cons(x,cons(y,k))))	(29)
sum^#(app(l,cons(x,cons(y,k))))	→	sum^#(cons(x,cons(y,k)))	(30)
plus^#(plus(0,y),ext)	→	plus^#(y,ext)	(31)
plus^#(plus(s(x),y),ext)	→	plus^#(s(plus(x,y)),ext)	(32)
plus^#(plus(s(x),y),ext)	→	plus^#(x,y)	(33)

The extended rules of the TRS

plus(plus(0,y),ext)	→	plus(y,ext)	(34)
plus(plus(s(x),y),ext)	→	plus(s(plus(x,y)),ext)	(35)

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 Dependency Graph Processor

The dependency pairs are split into 6 components.

The 1^st component contains the pair
quot^#(s(x),s(y)) → quot^#(minus(x,y),s(y)) (22)

1.1.1 AC Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals

[quot^#(x₁, x₂)] = 2 · x₁

[s(x₁)] = 2 + 3 · x₁

[minus(x₁, x₂)] = 1 + 2 · x₁

[plus(x₁, x₂)] = 0

[0] = 0

together with the usable rules

minus(minus(x,y),z) → minus(x,plus(y,z)) (3)

minus(s(x),s(y)) → minus(x,y) (2)

minus(x,0) → x (1)

(w.r.t. the implicit argument filter of the reduction pair), the pair
quot^#(s(x),s(y)) → quot^#(minus(x,y),s(y)) (22)
could be deleted.
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

minus^#(minus(x,y),z)	→	minus^#(x,plus(y,z))	(20)
minus^#(s(x),s(y))	→	minus^#(x,y)	(19)

1.1.2 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[plus(x₁, x₂)]	=	1 · x₁ + 1 · x₂
[s(x₁)]	=	3 + 1 · x₁
[0]	=	0
[minus^#(x₁, x₂)]	=	3 · x₁ + 2 · x₂
[minus(x₁, x₂)]	=	2 + 3 · x₁ + 3 · x₂

together with the usable rules

plus(plus(s(x),y),ext)	→	plus(s(plus(x,y)),ext)	(35)
plus(s(x),y)	→	s(plus(x,y))	(7)
plus(0,y)	→	y	(6)
plus(plus(0,y),ext)	→	plus(y,ext)	(34)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(15)
plus(x,y)	→	plus(y,x)	(14)

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

minus^#(minus(x,y),z)	→	minus^#(x,plus(y,z))	(20)
minus^#(s(x),s(y))	→	minus^#(x,y)	(19)

and the rules

minus(x,0)	→	x	(1)
minus(s(x),s(y))	→	minus(x,y)	(2)
minus(minus(x,y),z)	→	minus(x,plus(y,z))	(3)
quot(0,s(y))	→	0	(4)
quot(s(x),s(y))	→	s(quot(minus(x,y),s(y)))	(5)
plus(0,y)	→	y	(6)
app(nil,k)	→	k	(8)
app(l,nil)	→	l	(9)
app(cons(x,l),k)	→	cons(x,app(l,k))	(10)
sum(cons(x,nil))	→	cons(x,nil)	(11)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(12)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(13)
plus(plus(0,y),ext)	→	plus(y,ext)	(34)

could be deleted.

1.1.2.1 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

sum^#(app(l,cons(x,cons(y,k))))

→

sum^#(app(l,sum(cons(x,cons(y,k)))))

(28)

1.1.3 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

together with the usable rules

plus(plus(s(x),y),ext)	→	plus(s(plus(x,y)),ext)	(35)
app(cons(x,l),k)	→	cons(x,app(l,k))	(10)
plus(s(x),y)	→	s(plus(x,y))	(7)
sum(cons(x,nil))	→	cons(x,nil)	(11)
app(nil,k)	→	k	(8)
app(l,nil)	→	l	(9)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(13)
plus(0,y)	→	y	(6)
plus(plus(0,y),ext)	→	plus(y,ext)	(34)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(12)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(15)
plus(x,y)	→	plus(y,x)	(14)

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

minus(x,0)	→	x	(1)
minus(s(x),s(y))	→	minus(x,y)	(2)
minus(minus(x,y),z)	→	minus(x,plus(y,z))	(3)
quot(0,s(y))	→	0	(4)
quot(s(x),s(y))	→	s(quot(minus(x,y),s(y)))	(5)
plus(0,y)	→	y	(6)
plus(plus(0,y),ext)	→	plus(y,ext)	(34)

could be deleted.

1.1.3.1 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

together with the usable rules

plus(plus(s(x),y),ext)	→	plus(s(plus(x,y)),ext)	(35)
app(cons(x,l),k)	→	cons(x,app(l,k))	(10)
plus(s(x),y)	→	s(plus(x,y))	(7)
sum(cons(x,nil))	→	cons(x,nil)	(11)
app(nil,k)	→	k	(8)
app(l,nil)	→	l	(9)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(13)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(12)
plus(x,y)	→	plus(y,x)	(14)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(15)

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

app(nil,k)	→	k	(8)
app(l,nil)	→	l	(9)

could be deleted.

1.1.3.1.1 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

together with the usable rules

plus(plus(s(x),y),ext)	→	plus(s(plus(x,y)),ext)	(35)
app(cons(x,l),k)	→	cons(x,app(l,k))	(10)
plus(s(x),y)	→	s(plus(x,y))	(7)
sum(cons(x,nil))	→	cons(x,nil)	(11)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(13)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(12)
plus(x,y)	→	plus(y,x)	(14)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(15)

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

sum(cons(x,cons(y,l)))

→

sum(cons(plus(x,y),l))

(12)

could be deleted.

1.1.3.1.1.1 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[app(x₁, x₂)]	=	2 · x₁ + 2 · x₂
[cons(x₁, x₂)]	=	2 · x₁ + 1 · x₂
[sum(x₁)]	=	1 · x₁
[nil]	=	0
[sum^#(x₁)]	=	2 · x₁

together with the usable rules

app(cons(x,l),k)	→	cons(x,app(l,k))	(10)
sum(cons(x,nil))	→	cons(x,nil)	(11)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(13)

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

plus(plus(s(x),y),ext)	→	plus(s(plus(x,y)),ext)	(35)
plus(s(x),y)	→	s(plus(x,y))	(7)

could be deleted.

1.1.3.1.1.1.1 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[plus(x₁, x₂)]	=	1 · x₁ + 1 · x₂
[app(x₁, x₂)]	=	2 · x₁ + 2 · x₂
[cons(x₁, x₂)]	=	2 + 2 · x₁ + 1 · x₂
[sum(x₁)]	=	1 · x₁
[nil]	=	0
[sum^#(x₁)]	=	2 · x₁

together with the usable rules

app(cons(x,l),k)	→	cons(x,app(l,k))	(10)
sum(cons(x,nil))	→	cons(x,nil)	(11)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(13)
plus(x,y)	→	plus(y,x)	(14)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(15)

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

app(cons(x,l),k)

→

cons(x,app(l,k))

(10)

could be deleted.

1.1.3.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.

The 4^th component contains the pair

sum^#(cons(x,cons(y,l)))

→

sum^#(cons(plus(x,y),l))

(26)

1.1.4 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[plus(x₁, x₂)]	=	1 · x₁ + 1 · x₂
[s(x₁)]	=	3 + 1 · x₁
[0]	=	0
[sum^#(x₁)]	=	1 · x₁
[cons(x₁, x₂)]	=	2 + 1 · x₁ + 1 · x₂

together with the usable rules

plus(plus(s(x),y),ext)	→	plus(s(plus(x,y)),ext)	(35)
plus(s(x),y)	→	s(plus(x,y))	(7)
plus(0,y)	→	y	(6)
plus(plus(0,y),ext)	→	plus(y,ext)	(34)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(15)
plus(x,y)	→	plus(y,x)	(14)

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

sum^#(cons(x,cons(y,l)))

→

sum^#(cons(plus(x,y),l))

(26)

and the rules

minus(x,0)	→	x	(1)
minus(s(x),s(y))	→	minus(x,y)	(2)
minus(minus(x,y),z)	→	minus(x,plus(y,z))	(3)
quot(0,s(y))	→	0	(4)
quot(s(x),s(y))	→	s(quot(minus(x,y),s(y)))	(5)
plus(0,y)	→	y	(6)
app(nil,k)	→	k	(8)
app(l,nil)	→	l	(9)
app(cons(x,l),k)	→	cons(x,app(l,k))	(10)
sum(cons(x,nil))	→	cons(x,nil)	(11)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(12)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(13)
plus(plus(0,y),ext)	→	plus(y,ext)	(34)

could be deleted.

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.

The 5^th component contains the pair

plus^#(plus(0,y),ext)	→	plus^#(y,ext)	(31)
plus^#(s(x),y)	→	plus^#(x,y)	(24)
plus^#(plus(s(x),y),ext)	→	plus^#(s(plus(x,y)),ext)	(32)
plus^#(plus(s(x),y),ext)	→	plus^#(x,y)	(33)
plus^#(plus(x,y),z)	→	plus^#(y,z)	(18)
plus^#(x,y)	→	plus^#(y,x)	(16)
plus^#(plus(x,y),z)	→	plus^#(x,plus(y,z))	(17)

1.1.5 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(plus(s(x),y),ext)	→	plus(s(plus(x,y)),ext)	(35)
plus(s(x),y)	→	s(plus(x,y))	(7)
plus(0,y)	→	y	(6)
plus(plus(0,y),ext)	→	plus(y,ext)	(34)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(15)
plus(x,y)	→	plus(y,x)	(14)

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

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

→

plus^#(y,ext)

(31)

and the rules

minus(x,0)	→	x	(1)
minus(s(x),s(y))	→	minus(x,y)	(2)
minus(minus(x,y),z)	→	minus(x,plus(y,z))	(3)
quot(0,s(y))	→	0	(4)
quot(s(x),s(y))	→	s(quot(minus(x,y),s(y)))	(5)
plus(0,y)	→	y	(6)
app(nil,k)	→	k	(8)
app(l,nil)	→	l	(9)
app(cons(x,l),k)	→	cons(x,app(l,k))	(10)
sum(cons(x,nil))	→	cons(x,nil)	(11)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(12)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(13)
plus(plus(0,y),ext)	→	plus(y,ext)	(34)

could be deleted.

1.1.5.1 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 + 1 · x₁

together with the usable rules

plus(plus(s(x),y),ext)	→	plus(s(plus(x,y)),ext)	(35)
plus(s(x),y)	→	s(plus(x,y))	(7)
plus(x,y)	→	plus(y,x)	(14)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(15)

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

plus^#(s(x),y)	→	plus^#(x,y)	(24)
plus^#(plus(s(x),y),ext)	→	plus^#(x,y)	(33)

and no rules could be deleted.

1.1.5.1.1 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 + 1 · x₁ + 1 · x₂
[s(x₁)]	=	1 · x₁

together with the usable rules

plus(plus(s(x),y),ext)	→	plus(s(plus(x,y)),ext)	(35)
plus(s(x),y)	→	s(plus(x,y))	(7)
plus(x,y)	→	plus(y,x)	(14)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(15)

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

plus^#(plus(x,y),z)

→

plus^#(y,z)

(18)

and no rules could be deleted.

1.1.5.1.1.1 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(plus(s(x),y),ext)	→	plus(s(plus(x,y)),ext)	(35)
plus(s(x),y)	→	s(plus(x,y))	(7)
plus(plus(x,y),z)	→	plus(x,plus(y,z))	(15)
plus(x,y)	→	plus(y,x)	(14)

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

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

→

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

(32)

could be deleted.

1.1.5.1.1.1.1 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 6^th component contains the pair

app^#(cons(x,l),k)

→

app^#(l,k)

(25)

1.1.6 AC Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

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

app^#(cons(x,l),k)

→

app^#(l,k)

(25)

and the rules

minus(x,0)	→	x	(1)
minus(s(x),s(y))	→	minus(x,y)	(2)
minus(minus(x,y),z)	→	minus(x,plus(y,z))	(3)
quot(0,s(y))	→	0	(4)
quot(s(x),s(y))	→	s(quot(minus(x,y),s(y)))	(5)
plus(0,y)	→	y	(6)
plus(s(x),y)	→	s(plus(x,y))	(7)
app(nil,k)	→	k	(8)
app(l,nil)	→	l	(9)
app(cons(x,l),k)	→	cons(x,app(l,k))	(10)
sum(cons(x,nil))	→	cons(x,nil)	(11)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(12)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(13)
plus(plus(0,y),ext)	→	plus(y,ext)	(34)
plus(plus(s(x),y),ext)	→	plus(s(plus(x,y)),ext)	(35)

could be deleted.

1.1.6.1 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.

[quot^#(x₁, x₂)]	=	2 · x₁
[s(x₁)]	=	2 + 3 · x₁
[minus(x₁, x₂)]	=	1 + 2 · x₁
[plus(x₁, x₂)]	=	0
[0]	=	0