Certification Problem

Input (TPDB TRS_Standard/AG01/#3.17a)

The rewrite relation of the following TRS is considered.

app(nil,k)	→	k	(1)
app(l,nil)	→	l	(2)
app(cons(x,l),k)	→	cons(x,app(l,k))	(3)
sum(cons(x,nil))	→	cons(x,nil)	(4)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(5)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(6)
plus(0,y)	→	y	(7)
plus(s(x),y)	→	s(plus(x,y))	(8)
sum(plus(cons(0,x),cons(y,l)))	→	pred(sum(cons(s(x),cons(y,l))))	(9)
pred(cons(s(x),nil))	→	cons(x,nil)	(10)

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.

sum^#(app(l,cons(x,cons(y,k))))	→	app^#(l,sum(cons(x,cons(y,k))))	(11)
sum^#(cons(x,cons(y,l)))	→	sum^#(cons(plus(x,y),l))	(12)
plus^#(s(x),y)	→	plus^#(x,y)	(13)
sum^#(cons(x,cons(y,l)))	→	plus^#(x,y)	(14)
sum^#(plus(cons(0,x),cons(y,l)))	→	sum^#(cons(s(x),cons(y,l)))	(15)
sum^#(app(l,cons(x,cons(y,k))))	→	sum^#(cons(x,cons(y,k)))	(16)
sum^#(app(l,cons(x,cons(y,k))))	→	sum^#(app(l,sum(cons(x,cons(y,k)))))	(17)
app^#(cons(x,l),k)	→	app^#(l,k)	(18)
sum^#(plus(cons(0,x),cons(y,l)))	→	pred^#(sum(cons(s(x),cons(y,l))))	(19)

1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.

The 1^st component contains the pair

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

→

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

(17)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[s(x₁)]	=	x₁ + 1
[plus^#(x₁, x₂)]	=	0
[pred(x₁)]	=	x₁ + 0
[sum(x₁)]	=	3
[0]	=	1
[nil]	=	1
[app^#(x₁, x₂)]	=	0
[plus(x₁, x₂)]	=	x₁ + x₂ + 14235
[pred^#(x₁)]	=	0
[cons(x₁, x₂)]	=	x₂ + 2
[sum^#(x₁)]	=	x₁ + 0
[app(x₁, x₂)]	=	x₁ + x₂ + 17221

together with the usable rules

sum(cons(x,nil))	→	cons(x,nil)	(4)
app(nil,k)	→	k	(1)
app(cons(x,l),k)	→	cons(x,app(l,k))	(3)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(5)
pred(cons(s(x),nil))	→	cons(x,nil)	(10)
sum(plus(cons(0,x),cons(y,l)))	→	pred(sum(cons(s(x),cons(y,l))))	(9)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(6)
app(l,nil)	→	l	(2)

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

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

→

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

(17)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.

The 2^nd component contains the pair

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

→

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

(12)

1.1.2 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[s(x₁)]	=	x₁ + 1
[plus^#(x₁, x₂)]	=	0
[pred(x₁)]	=	x₁ + 0
[sum(x₁)]	=	2
[0]	=	38551
[nil]	=	1
[app^#(x₁, x₂)]	=	0
[plus(x₁, x₂)]	=	x₁ + x₂ + 1
[pred^#(x₁)]	=	0
[cons(x₁, x₂)]	=	x₂ + 1
[sum^#(x₁)]	=	x₁ + 0
[app(x₁, x₂)]	=	x₁ + x₂ + 36466

together with the usable rules

sum(cons(x,nil))	→	cons(x,nil)	(4)
app(nil,k)	→	k	(1)
app(cons(x,l),k)	→	cons(x,app(l,k))	(3)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(5)
pred(cons(s(x),nil))	→	cons(x,nil)	(10)
sum(plus(cons(0,x),cons(y,l)))	→	pred(sum(cons(s(x),cons(y,l))))	(9)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(6)
app(l,nil)	→	l	(2)

(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))

(12)

could be deleted.

1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.

The 3^rd component contains the pair

plus^#(s(x),y)

→

plus^#(x,y)

(13)

1.1.3 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[s(x₁)]	=	x₁ + 1
[plus^#(x₁, x₂)]	=	x₁ + 0
[pred(x₁)]	=	x₁ + 0
[sum(x₁)]	=	16204
[0]	=	1
[nil]	=	16203
[app^#(x₁, x₂)]	=	0
[plus(x₁, x₂)]	=	x₁ + x₂ + 1
[pred^#(x₁)]	=	0
[cons(x₁, x₂)]	=	x₂ + 1
[sum^#(x₁)]	=	x₁ + 0
[app(x₁, x₂)]	=	x₁ + x₂ + 29483

together with the usable rules

sum(cons(x,nil))	→	cons(x,nil)	(4)
app(nil,k)	→	k	(1)
app(cons(x,l),k)	→	cons(x,app(l,k))	(3)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(5)
pred(cons(s(x),nil))	→	cons(x,nil)	(10)
sum(plus(cons(0,x),cons(y,l)))	→	pred(sum(cons(s(x),cons(y,l))))	(9)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(6)
app(l,nil)	→	l	(2)

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

plus^#(s(x),y)

→

plus^#(x,y)

(13)

could be deleted.

1.1.3.1 Dependency Graph Processor

The dependency pairs are split into 0 components.

The 4^th component contains the pair

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

→

app^#(l,k)

(18)

1.1.4 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[s(x₁)]	=	x₁ + 1
[plus^#(x₁, x₂)]	=	0
[pred(x₁)]	=	x₁ + 0
[sum(x₁)]	=	23975
[0]	=	1
[nil]	=	23974
[app^#(x₁, x₂)]	=	x₁ + 0
[plus(x₁, x₂)]	=	x₁ + x₂ + 1
[pred^#(x₁)]	=	0
[cons(x₁, x₂)]	=	x₂ + 1
[sum^#(x₁)]	=	x₁ + 0
[app(x₁, x₂)]	=	x₁ + x₂ + 46572

together with the usable rules

sum(cons(x,nil))	→	cons(x,nil)	(4)
app(nil,k)	→	k	(1)
app(cons(x,l),k)	→	cons(x,app(l,k))	(3)
sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(5)
pred(cons(s(x),nil))	→	cons(x,nil)	(10)
sum(plus(cons(0,x),cons(y,l)))	→	pred(sum(cons(s(x),cons(y,l))))	(9)
sum(app(l,cons(x,cons(y,k))))	→	sum(app(l,sum(cons(x,cons(y,k)))))	(6)
app(l,nil)	→	l	(2)

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

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

→

app^#(l,k)

(18)

could be deleted.

1.1.4.1 Dependency Graph Processor

The dependency pairs are split into 0 components.