Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/013)

The rewrite relation of the following TRS is considered.

app(app(append,nil),ys) → ys (1)
app(app(append,app(app(cons,x),xs)),ys) → app(app(cons,x),app(app(append,xs),ys)) (2)
app(app(flatwith,f),app(leaf,x)) → app(app(cons,app(f,x)),nil) (3)
app(app(flatwith,f),app(node,xs)) → app(app(flatwithsub,f),xs) (4)
app(app(flatwithsub,f),nil) → nil (5)
app(app(flatwithsub,f),app(app(cons,x),xs)) → app(app(append,app(app(flatwith,f),x)),app(app(flatwithsub,f),xs)) (6)

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.

append is mapped to append, append1(x1), append2(x1, x2)
nil is mapped to nil
cons is mapped to cons, cons1(x1), cons2(x1, x2)
flatwith is mapped to flatwith, flatwith1(x1), flatwith2(x1, x2)
leaf is mapped to leaf, leaf1(x1)
node is mapped to node, node1(x1)
flatwithsub is mapped to flatwithsub, flatwithsub1(x1), flatwithsub2(x1, x2)


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
append2(nil,ys) → ys (17)
append2(cons2(x,xs),ys) → cons2(x,append2(xs,ys)) (18)
flatwith2(f,leaf1(x)) → cons2(app(f,x),nil) (19)
flatwith2(f,node1(xs)) → flatwithsub2(f,xs) (20)
flatwithsub2(f,nil) → nil (21)
flatwithsub2(f,cons2(x,xs)) → append2(flatwith2(f,x),flatwithsub2(f,xs)) (22)
app(append,y1) → append1(y1) (7)
app(append1(x0),y1) → append2(x0,y1) (8)
app(cons,y1) → cons1(y1) (9)
app(cons1(x0),y1) → cons2(x0,y1) (10)
app(flatwith,y1) → flatwith1(y1) (11)
app(flatwith1(x0),y1) → flatwith2(x0,y1) (12)
app(leaf,y1) → leaf1(y1) (13)
app(node,y1) → node1(y1) (14)
app(flatwithsub,y1) → flatwithsub1(y1) (15)
app(flatwithsub1(x0),y1) → flatwithsub2(x0,y1) (16)

1.1 Rule Removal

Using the
prec(append2) = 2 stat(append2) = mul
prec(nil) = 3 stat(nil) = mul
prec(cons2) = 1 stat(cons2) = mul
prec(flatwith2) = 4 stat(flatwith2) = mul
prec(leaf1) = 3 stat(leaf1) = mul
prec(app) = 4 stat(app) = mul
prec(node1) = 0 stat(node1) = mul
prec(flatwithsub2) = 4 stat(flatwithsub2) = mul
prec(append) = 5 stat(append) = mul
prec(cons) = 6 stat(cons) = mul
prec(cons1) = 0 stat(cons1) = mul
prec(flatwith) = 7 stat(flatwith) = mul
prec(flatwith1) = 4 stat(flatwith1) = mul
prec(leaf) = 3 stat(leaf) = mul
prec(node) = 8 stat(node) = mul
prec(flatwithsub) = 0 stat(flatwithsub) = mul
prec(flatwithsub1) = 0 stat(flatwithsub1) = mul

π(append2) = [1,2]
π(nil) = []
π(cons2) = [1,2]
π(flatwith2) = [1,2]
π(leaf1) = [1]
π(app) = [1,2]
π(node1) = [1]
π(flatwithsub2) = [1,2]
π(append) = []
π(append1) = 1
π(cons) = []
π(cons1) = [1]
π(flatwith) = []
π(flatwith1) = [1]
π(leaf) = []
π(node) = []
π(flatwithsub) = []
π(flatwithsub1) = [1]

all of the following rules can be deleted.
append2(nil,ys) → ys (17)
append2(cons2(x,xs),ys) → cons2(x,append2(xs,ys)) (18)
flatwith2(f,leaf1(x)) → cons2(app(f,x),nil) (19)
flatwith2(f,node1(xs)) → flatwithsub2(f,xs) (20)
flatwithsub2(f,nil) → nil (21)
flatwithsub2(f,cons2(x,xs)) → append2(flatwith2(f,x),flatwithsub2(f,xs)) (22)
app(append,y1) → append1(y1) (7)
app(append1(x0),y1) → append2(x0,y1) (8)
app(cons,y1) → cons1(y1) (9)
app(cons1(x0),y1) → cons2(x0,y1) (10)
app(flatwith,y1) → flatwith1(y1) (11)
app(flatwith1(x0),y1) → flatwith2(x0,y1) (12)
app(leaf,y1) → leaf1(y1) (13)
app(node,y1) → node1(y1) (14)
app(flatwithsub,y1) → flatwithsub1(y1) (15)
app(flatwithsub1(x0),y1) → flatwithsub2(x0,y1) (16)

1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.