Certification Problem

Input (TPDB TRS_Standard/Applicative_05/Hamming)

The rewrite relation of the following TRS is considered.

app(app(app(if,true),xs),ys) → xs (1)
app(app(app(if,false),xs),ys) → ys (2)
app(app(lt,app(s,x)),app(s,y)) → app(app(lt,x),y) (3)
app(app(lt,0),app(s,y)) → true (4)
app(app(lt,y),0) → false (5)
app(app(eq,x),x) → true (6)
app(app(eq,app(s,x)),0) → false (7)
app(app(eq,0),app(s,x)) → false (8)
app(app(merge,xs),nil) → xs (9)
app(app(merge,nil),ys) → ys (10)
app(app(merge,app(app(cons,x),xs)),app(app(cons,y),ys)) → app(app(app(if,app(app(lt,x),y)),app(app(cons,x),app(app(merge,xs),app(app(cons,y),ys)))),app(app(app(if,app(app(eq,x),y)),app(app(cons,x),app(app(merge,xs),ys))),app(app(cons,y),app(app(merge,app(app(cons,x),xs)),ys)))) (11)
app(app(map,f),nil) → nil (12)
app(app(map,f),app(app(cons,x),xs)) → app(app(cons,app(f,x)),app(app(map,f),xs)) (13)
app(app(mult,0),x) → 0 (14)
app(app(mult,app(s,x)),y) → app(app(plus,y),app(app(mult,x),y)) (15)
app(app(plus,0),x) → 0 (16)
app(app(plus,app(s,x)),y) → app(s,app(app(plus,x),y)) (17)
list1 → app(app(map,app(mult,app(s,app(s,0)))),hamming) (18)
list2 → app(app(map,app(mult,app(s,app(s,app(s,0))))),hamming) (19)
list3 → app(app(map,app(mult,app(s,app(s,app(s,app(s,app(s,0))))))),hamming) (20)
hamming → app(app(cons,app(s,0)),app(app(merge,list1),app(app(merge,list2),list3))) (21)

Property / Task

Prove or disprove termination.

Answer / Result

No.

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.

if is mapped to if, if1(x1), if2(x1, x2), if3(x1, x2, x3)
true is mapped to true
false is mapped to false
lt is mapped to lt, lt1(x1), lt2(x1, x2)
s is mapped to s, s1(x1)
0 is mapped to 0
eq is mapped to eq, eq1(x1), eq2(x1, x2)
merge is mapped to merge, merge1(x1), merge2(x1, x2)
nil is mapped to nil
cons is mapped to cons, cons1(x1), cons2(x1, x2)
map is mapped to map, map1(x1), map2(x1, x2)
mult is mapped to mult, mult1(x1), mult2(x1, x2)
plus is mapped to plus, plus1(x1), plus2(x1, x2)
list1 is mapped to list1
hamming is mapped to hamming
list2 is mapped to list2
list3 is mapped to list3


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
if3(true,xs,ys) → xs (40)
if3(false,xs,ys) → ys (41)
lt2(s1(x),s1(y)) → lt2(x,y) (42)
lt2(0,s1(y)) → true (43)
lt2(y,0) → false (44)
eq2(x,x) → true (45)
eq2(s1(x),0) → false (46)
eq2(0,s1(x)) → false (47)
merge2(xs,nil) → xs (48)
merge2(nil,ys) → ys (49)
merge2(cons2(x,xs),cons2(y,ys)) → if3(lt2(x,y),cons2(x,merge2(xs,cons2(y,ys))),if3(eq2(x,y),cons2(x,merge2(xs,ys)),cons2(y,merge2(cons2(x,xs),ys)))) (50)
map2(f,nil) → nil (51)
map2(f,cons2(x,xs)) → cons2(app(f,x),map2(f,xs)) (52)
mult2(0,x) → 0 (53)
mult2(s1(x),y) → plus2(y,mult2(x,y)) (54)
plus2(0,x) → 0 (55)
plus2(s1(x),y) → s1(plus2(x,y)) (56)
list1 → map2(mult1(s1(s1(0))),hamming) (57)
list2 → map2(mult1(s1(s1(s1(0)))),hamming) (58)
list3 → map2(mult1(s1(s1(s1(s1(s1(0)))))),hamming) (59)
hamming → cons2(s1(0),merge2(list1,merge2(list2,list3))) (60)
app(if,y1) → if1(y1) (22)
app(if1(x0),y1) → if2(x0,y1) (23)
app(if2(x0,x1),y1) → if3(x0,x1,y1) (24)
app(lt,y1) → lt1(y1) (25)
app(lt1(x0),y1) → lt2(x0,y1) (26)
app(s,y1) → s1(y1) (27)
app(eq,y1) → eq1(y1) (28)
app(eq1(x0),y1) → eq2(x0,y1) (29)
app(merge,y1) → merge1(y1) (30)
app(merge1(x0),y1) → merge2(x0,y1) (31)
app(cons,y1) → cons1(y1) (32)
app(cons1(x0),y1) → cons2(x0,y1) (33)
app(map,y1) → map1(y1) (34)
app(map1(x0),y1) → map2(x0,y1) (35)
app(mult,y1) → mult1(y1) (36)
app(mult1(x0),y1) → mult2(x0,y1) (37)
app(plus,y1) → plus1(y1) (38)
app(plus1(x0),y1) → plus2(x0,y1) (39)

1.1 Innermost Lhss Increase

We add the following left hand sides to the innermost strategy.
if3(true,x0,x1)
if3(false,x0,x1)
lt2(s1(x0),s1(x1))
lt2(0,s1(x0))
lt2(x0,0)
eq2(x0,x0)
eq2(s1(x0),0)
eq2(0,s1(x0))
merge2(x0,nil)
merge2(nil,x0)
merge2(cons2(x0,x1),cons2(x2,x3))
map2(x0,nil)
map2(x0,cons2(x1,x2))
mult2(0,x0)
mult2(s1(x0),x1)
plus2(0,x0)
plus2(s1(x0),x1)
list1
list2
list3
hamming
app(if,x0)
app(if1(x0),x1)
app(if2(x0,x1),x2)
app(lt,x0)
app(lt1(x0),x1)
app(s,x0)
app(eq,x0)
app(eq1(x0),x1)
app(merge,x0)
app(merge1(x0),x1)
app(cons,x0)
app(cons1(x0),x1)
app(map,x0)
app(map1(x0),x1)
app(mult,x0)
app(mult1(x0),x1)
app(plus,x0)
app(plus1(x0),x1)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
lt2#(s1(x),s1(y)) → lt2#(x,y) (61)
merge2#(cons2(x,xs),cons2(y,ys)) → if3#(lt2(x,y),cons2(x,merge2(xs,cons2(y,ys))),if3(eq2(x,y),cons2(x,merge2(xs,ys)),cons2(y,merge2(cons2(x,xs),ys)))) (62)
merge2#(cons2(x,xs),cons2(y,ys)) → lt2#(x,y) (63)
merge2#(cons2(x,xs),cons2(y,ys)) → merge2#(xs,cons2(y,ys)) (64)
merge2#(cons2(x,xs),cons2(y,ys)) → if3#(eq2(x,y),cons2(x,merge2(xs,ys)),cons2(y,merge2(cons2(x,xs),ys))) (65)
merge2#(cons2(x,xs),cons2(y,ys)) → eq2#(x,y) (66)
merge2#(cons2(x,xs),cons2(y,ys)) → merge2#(xs,ys) (67)
merge2#(cons2(x,xs),cons2(y,ys)) → merge2#(cons2(x,xs),ys) (68)
map2#(f,cons2(x,xs)) → app#(f,x) (69)
map2#(f,cons2(x,xs)) → map2#(f,xs) (70)
mult2#(s1(x),y) → plus2#(y,mult2(x,y)) (71)
mult2#(s1(x),y) → mult2#(x,y) (72)
plus2#(s1(x),y) → plus2#(x,y) (73)
list1# → map2#(mult1(s1(s1(0))),hamming) (74)
list1# → hamming# (75)
list2# → map2#(mult1(s1(s1(s1(0)))),hamming) (76)
list2# → hamming# (77)
list3# → map2#(mult1(s1(s1(s1(s1(s1(0)))))),hamming) (78)
list3# → hamming# (79)
hamming# → merge2#(list1,merge2(list2,list3)) (80)
hamming# → list1# (81)
hamming# → merge2#(list2,list3) (82)
hamming# → list2# (83)
hamming# → list3# (84)
app#(if2(x0,x1),y1) → if3#(x0,x1,y1) (85)
app#(lt1(x0),y1) → lt2#(x0,y1) (86)
app#(eq1(x0),y1) → eq2#(x0,y1) (87)
app#(merge1(x0),y1) → merge2#(x0,y1) (88)
app#(map1(x0),y1) → map2#(x0,y1) (89)
app#(mult1(x0),y1) → mult2#(x0,y1) (90)
app#(plus1(x0),y1) → plus2#(x0,y1) (91)
It remains to prove infiniteness of the resulting DP problem.

1.1.1.1 Pair and Rule Removal

Some pairs and rules have been removed and it remains to prove infiniteness of the remaing problem. The following pairs have been deleted.
lt2#(s1(x),s1(y)) → lt2#(x,y) (61)
merge2#(cons2(x,xs),cons2(y,ys)) → if3#(lt2(x,y),cons2(x,merge2(xs,cons2(y,ys))),if3(eq2(x,y),cons2(x,merge2(xs,ys)),cons2(y,merge2(cons2(x,xs),ys)))) (62)
merge2#(cons2(x,xs),cons2(y,ys)) → lt2#(x,y) (63)
merge2#(cons2(x,xs),cons2(y,ys)) → merge2#(xs,cons2(y,ys)) (64)
merge2#(cons2(x,xs),cons2(y,ys)) → if3#(eq2(x,y),cons2(x,merge2(xs,ys)),cons2(y,merge2(cons2(x,xs),ys))) (65)
merge2#(cons2(x,xs),cons2(y,ys)) → eq2#(x,y) (66)
merge2#(cons2(x,xs),cons2(y,ys)) → merge2#(xs,ys) (67)
merge2#(cons2(x,xs),cons2(y,ys)) → merge2#(cons2(x,xs),ys) (68)
map2#(f,cons2(x,xs)) → app#(f,x) (69)
map2#(f,cons2(x,xs)) → map2#(f,xs) (70)
mult2#(s1(x),y) → plus2#(y,mult2(x,y)) (71)
mult2#(s1(x),y) → mult2#(x,y) (72)
plus2#(s1(x),y) → plus2#(x,y) (73)
list1# → map2#(mult1(s1(s1(0))),hamming) (74)
list2# → map2#(mult1(s1(s1(s1(0)))),hamming) (76)
list3# → map2#(mult1(s1(s1(s1(s1(s1(0)))))),hamming) (78)
hamming# → merge2#(list1,merge2(list2,list3)) (80)
hamming# → merge2#(list2,list3) (82)
app#(if2(x0,x1),y1) → if3#(x0,x1,y1) (85)
app#(lt1(x0),y1) → lt2#(x0,y1) (86)
app#(eq1(x0),y1) → eq2#(x0,y1) (87)
app#(merge1(x0),y1) → merge2#(x0,y1) (88)
app#(map1(x0),y1) → map2#(x0,y1) (89)
app#(mult1(x0),y1) → mult2#(x0,y1) (90)
app#(plus1(x0),y1) → plus2#(x0,y1) (91)
and the following rules have been deleted.

1.1.1.1.1 Pair and Rule Removal

Some pairs and rules have been removed and it remains to prove infiniteness of the remaing problem. The following pairs have been deleted. and the following rules have been deleted.
if3(true,xs,ys) → xs (40)
if3(false,xs,ys) → ys (41)
lt2(s1(x),s1(y)) → lt2(x,y) (42)
lt2(0,s1(y)) → true (43)
lt2(y,0) → false (44)
eq2(x,x) → true (45)
eq2(s1(x),0) → false (46)
eq2(0,s1(x)) → false (47)
merge2(xs,nil) → xs (48)
merge2(nil,ys) → ys (49)
merge2(cons2(x,xs),cons2(y,ys)) → if3(lt2(x,y),cons2(x,merge2(xs,cons2(y,ys))),if3(eq2(x,y),cons2(x,merge2(xs,ys)),cons2(y,merge2(cons2(x,xs),ys)))) (50)
map2(f,nil) → nil (51)
map2(f,cons2(x,xs)) → cons2(app(f,x),map2(f,xs)) (52)
mult2(0,x) → 0 (53)
mult2(s1(x),y) → plus2(y,mult2(x,y)) (54)
plus2(0,x) → 0 (55)
plus2(s1(x),y) → s1(plus2(x,y)) (56)
list1 → map2(mult1(s1(s1(0))),hamming) (57)
list2 → map2(mult1(s1(s1(s1(0)))),hamming) (58)
list3 → map2(mult1(s1(s1(s1(s1(s1(0)))))),hamming) (59)
hamming → cons2(s1(0),merge2(list1,merge2(list2,list3))) (60)
app(if,y1) → if1(y1) (22)
app(if1(x0),y1) → if2(x0,y1) (23)
app(if2(x0,x1),y1) → if3(x0,x1,y1) (24)
app(lt,y1) → lt1(y1) (25)
app(lt1(x0),y1) → lt2(x0,y1) (26)
app(s,y1) → s1(y1) (27)
app(eq,y1) → eq1(y1) (28)
app(eq1(x0),y1) → eq2(x0,y1) (29)
app(merge,y1) → merge1(y1) (30)
app(merge1(x0),y1) → merge2(x0,y1) (31)
app(cons,y1) → cons1(y1) (32)
app(cons1(x0),y1) → cons2(x0,y1) (33)
app(map,y1) → map1(y1) (34)
app(map1(x0),y1) → map2(x0,y1) (35)
app(mult,y1) → mult1(y1) (36)
app(mult1(x0),y1) → mult2(x0,y1) (37)
app(plus,y1) → plus1(y1) (38)
app(plus1(x0),y1) → plus2(x0,y1) (39)

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 Loop

The following loop proves infiniteness of the DP problem.

t0 = list1#
→P hamming#
→P list1#
= t2
where t2 = t0