Certification Problem

Input (TPDB TRS_Standard/Applicative_first_order_05/#3.45)

The rewrite relation of the following TRS is considered.

app(f,app(s,x))	→	app(f,x)	(1)
app(g,app(app(cons,0),y))	→	app(g,y)	(2)
app(g,app(app(cons,app(s,x)),y))	→	app(s,x)	(3)
app(h,app(app(cons,x),y))	→	app(h,app(g,app(app(cons,x),y)))	(4)
app(app(map,fun),nil)	→	nil	(5)
app(app(map,fun),app(app(cons,x),xs))	→	app(app(cons,app(fun,x)),app(app(map,fun),xs))	(6)
app(app(filter,fun),nil)	→	nil	(7)
app(app(filter,fun),app(app(cons,x),xs))	→	app(app(app(app(filter2,app(fun,x)),fun),x),xs)	(8)
app(app(app(app(filter2,true),fun),x),xs)	→	app(app(cons,x),app(app(filter,fun),xs))	(9)
app(app(app(app(filter2,false),fun),x),xs)	→	app(app(filter,fun),xs)	(10)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Uncurrying

We uncurry the binary symbol app in combination with the following symbol map which also determines the applicative arities of these symbols.

f	is mapped to	f,	f1(x₁)
s	is mapped to	s,	s1(x₁)
g	is mapped to	g,	g1(x₁)
cons	is mapped to	cons,	cons1(x₁),	cons2(x₁, x₂)
0	is mapped to	0
h	is mapped to	h,	h1(x₁)
map	is mapped to	map,	map1(x₁),	map2(x₁, x₂)
nil	is mapped to	nil
filter	is mapped to	filter,	filter1(x₁),	filter3(x₁, x₂)
filter2	is mapped to	filter2,	filter21(x₁),	filter22(x₁, x₂),	filter23(x₁, x₂, x₃),	filter24(x₁,...,x₄)
true	is mapped to	true
false	is mapped to	false

There are no uncurry rules.
No rules have to be added for the eta-expansion.

Uncurrying the rules and adding the uncurrying rules yields the new set of rules

f1(s1(x))	→	f1(x)	(25)
g1(cons2(0,y))	→	g1(y)	(26)
g1(cons2(s1(x),y))	→	s1(x)	(27)
h1(cons2(x,y))	→	h1(g1(cons2(x,y)))	(28)
map2(fun,nil)	→	nil	(29)
map2(fun,cons2(x,xs))	→	cons2(app(fun,x),map2(fun,xs))	(30)
filter3(fun,nil)	→	nil	(31)
filter3(fun,cons2(x,xs))	→	filter24(app(fun,x),fun,x,xs)	(32)
filter24(true,fun,x,xs)	→	cons2(x,filter3(fun,xs))	(33)
filter24(false,fun,x,xs)	→	filter3(fun,xs)	(34)
app(f,y1)	→	f1(y1)	(11)
app(s,y1)	→	s1(y1)	(12)
app(g,y1)	→	g1(y1)	(13)
app(cons,y1)	→	cons1(y1)	(14)
app(cons1(x0),y1)	→	cons2(x0,y1)	(15)
app(h,y1)	→	h1(y1)	(16)
app(map,y1)	→	map1(y1)	(17)
app(map1(x0),y1)	→	map2(x0,y1)	(18)
app(filter,y1)	→	filter1(y1)	(19)
app(filter1(x0),y1)	→	filter3(x0,y1)	(20)
app(filter2,y1)	→	filter21(y1)	(21)
app(filter21(x0),y1)	→	filter22(x0,y1)	(22)
app(filter22(x0,x1),y1)	→	filter23(x0,x1,y1)	(23)
app(filter23(x0,x1,x2),y1)	→	filter24(x0,x1,x2,y1)	(24)

1.1 Rule Removal

Using the

prec(f1)	=	0	stat(f1)	=	mul
prec(cons2)	=	1	stat(cons2)	=	mul
prec(0)	=	2	stat(0)	=	mul
prec(map2)	=	6	stat(map2)	=	lex
prec(nil)	=	7	stat(nil)	=	mul
prec(app)	=	6	stat(app)	=	lex
prec(filter3)	=	6	stat(filter3)	=	lex
prec(filter24)	=	6	stat(filter24)	=	lex
prec(true)	=	8	stat(true)	=	mul
prec(false)	=	9	stat(false)	=	mul
prec(f)	=	10	stat(f)	=	mul
prec(s)	=	11	stat(s)	=	mul
prec(g)	=	12	stat(g)	=	mul
prec(cons)	=	13	stat(cons)	=	mul
prec(cons1)	=	3	stat(cons1)	=	mul
prec(h)	=	14	stat(h)	=	mul
prec(map)	=	15	stat(map)	=	mul
prec(filter)	=	16	stat(filter)	=	mul
prec(filter2)	=	17	stat(filter2)	=	mul
prec(filter21)	=	5	stat(filter21)	=	mul
prec(filter22)	=	4	stat(filter22)	=	mul
prec(filter23)	=	6	stat(filter23)	=	lex

π(f1)	=	[1]
π(s1)	=	1
π(g1)	=	1
π(cons2)	=	[1,2]
π(0)	=	[]
π(h1)	=	1
π(map2)	=	[2,1]
π(nil)	=	[]
π(app)	=	[2,1]
π(filter3)	=	[2,1]
π(filter24)	=	[4,2,1,3]
π(true)	=	[]
π(false)	=	[]
π(f)	=	[]
π(s)	=	[]
π(g)	=	[]
π(cons)	=	[]
π(cons1)	=	[1]
π(h)	=	[]
π(map)	=	[]
π(map1)	=	1
π(filter)	=	[]
π(filter1)	=	1
π(filter2)	=	[]
π(filter21)	=	[1]
π(filter22)	=	[1,2]
π(filter23)	=	[3,1,2]

all of the following rules can be deleted.

g1(cons2(0,y))	→	g1(y)	(26)
g1(cons2(s1(x),y))	→	s1(x)	(27)
map2(fun,nil)	→	nil	(29)
map2(fun,cons2(x,xs))	→	cons2(app(fun,x),map2(fun,xs))	(30)
filter3(fun,nil)	→	nil	(31)
filter3(fun,cons2(x,xs))	→	filter24(app(fun,x),fun,x,xs)	(32)
filter24(true,fun,x,xs)	→	cons2(x,filter3(fun,xs))	(33)
filter24(false,fun,x,xs)	→	filter3(fun,xs)	(34)
app(f,y1)	→	f1(y1)	(11)
app(s,y1)	→	s1(y1)	(12)
app(g,y1)	→	g1(y1)	(13)
app(cons,y1)	→	cons1(y1)	(14)
app(cons1(x0),y1)	→	cons2(x0,y1)	(15)
app(h,y1)	→	h1(y1)	(16)
app(map,y1)	→	map1(y1)	(17)
app(filter,y1)	→	filter1(y1)	(19)
app(filter2,y1)	→	filter21(y1)	(21)
app(filter21(x0),y1)	→	filter22(x0,y1)	(22)
app(filter22(x0,x1),y1)	→	filter23(x0,x1,y1)	(23)
app(filter23(x0,x1,x2),y1)	→	filter24(x0,x1,x2,y1)	(24)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals

[f1(x₁)]	=	2 · x₁
[s1(x₁)]	=	2 + 2 · x₁
[h1(x₁)]	=	2 · x₁
[cons2(x₁, x₂)]	=	1 · x₁ + 1 · x₂
[g1(x₁)]	=	1 · x₁
[app(x₁, x₂)]	=	2 + 1 · x₁ + 2 · x₂
[map1(x₁)]	=	2 + 2 · x₁
[map2(x₁, x₂)]	=	2 · x₁ + 2 · x₂
[filter1(x₁)]	=	2 + 2 · x₁
[filter3(x₁, x₂)]	=	2 · x₁ + 2 · x₂

all of the following rules can be deleted.

f1(s1(x))	→	f1(x)	(25)
app(map1(x0),y1)	→	map2(x0,y1)	(18)
app(filter1(x0),y1)	→	filter3(x0,y1)	(20)

1.1.1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

Hence, it suffices to show innermost termination in the following.

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

h1^#(cons2(x,y))

→

h1^#(g1(cons2(x,y)))

(35)

1.1.1.1.1.1 Usable Rules Processor

We restrict the rewrite rules to the following usable rules of the DP problem.

There are no rules.

1.1.1.1.1.1.1 Innermost Lhss Removal Processor

We restrict the innermost strategy to the following left hand sides.

There are no lhss.

1.1.1.1.1.1.1.1 Reduction Pair Processor

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions

prec(cons2)	=	1		weight(cons2)	=	2
prec(g1)	=	0		weight(g1)	=	1

in combination with the following argument filter

π(h1^#)	=	1
π(cons2)	=	[]
π(g1)	=	[]

the pair

h1^#(cons2(x,y))

→

h1^#(g1(cons2(x,y)))

(35)

could be deleted.

1.1.1.1.1.1.1.1.1 P is empty

There are no pairs anymore.