Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/012)

The rewrite relation of the following TRS is considered.

app(app(and,true),true) → true (1)
app(app(and,x),false) → false (2)
app(app(and,false),y) → false (3)
app(app(or,true),y) → true (4)
app(app(or,x),true) → true (5)
app(app(or,false),false) → false (6)
app(app(forall,p),nil) → true (7)
app(app(forall,p),app(app(cons,x),xs)) → app(app(and,app(p,x)),app(app(forall,p),xs)) (8)
app(app(forsome,p),nil) → false (9)
app(app(forsome,p),app(app(cons,x),xs)) → app(app(or,app(p,x)),app(app(forsome,p),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.

and is mapped to and, and1(x1), and2(x1, x2)
true is mapped to true
false is mapped to false
or is mapped to or, or1(x1), or2(x1, x2)
forall is mapped to forall, forall1(x1), forall2(x1, x2)
nil is mapped to nil
cons is mapped to cons, cons1(x1), cons2(x1, x2)
forsome is mapped to forsome, forsome1(x1), forsome2(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
and2(true,true) → true (21)
and2(x,false) → false (22)
and2(false,y) → false (23)
or2(true,y) → true (24)
or2(x,true) → true (25)
or2(false,false) → false (26)
forall2(p,nil) → true (27)
forall2(p,cons2(x,xs)) → and2(app(p,x),forall2(p,xs)) (28)
forsome2(p,nil) → false (29)
forsome2(p,cons2(x,xs)) → or2(app(p,x),forsome2(p,xs)) (30)
app(and,y1) → and1(y1) (11)
app(and1(x0),y1) → and2(x0,y1) (12)
app(or,y1) → or1(y1) (13)
app(or1(x0),y1) → or2(x0,y1) (14)
app(forall,y1) → forall1(y1) (15)
app(forall1(x0),y1) → forall2(x0,y1) (16)
app(cons,y1) → cons1(y1) (17)
app(cons1(x0),y1) → cons2(x0,y1) (18)
app(forsome,y1) → forsome1(y1) (19)
app(forsome1(x0),y1) → forsome2(x0,y1) (20)

1.1 Rule Removal

Using the
prec(and2) = 0 stat(and2) = mul
prec(true) = 1 stat(true) = mul
prec(false) = 0 stat(false) = mul
prec(or2) = 1 stat(or2) = mul
prec(forall2) = 4 stat(forall2) = mul
prec(nil) = 1 stat(nil) = mul
prec(cons2) = 2 stat(cons2) = lex
prec(app) = 4 stat(app) = mul
prec(forsome2) = 4 stat(forsome2) = mul
prec(and) = 5 stat(and) = mul
prec(and1) = 4 stat(and1) = mul
prec(or) = 6 stat(or) = mul
prec(or1) = 4 stat(or1) = mul
prec(forall) = 7 stat(forall) = mul
prec(forall1) = 4 stat(forall1) = mul
prec(cons) = 8 stat(cons) = mul
prec(cons1) = 4 stat(cons1) = mul
prec(forsome) = 9 stat(forsome) = mul
prec(forsome1) = 3 stat(forsome1) = lex

π(and2) = [1,2]
π(true) = []
π(false) = []
π(or2) = [1,2]
π(forall2) = [1,2]
π(nil) = []
π(cons2) = [2,1]
π(app) = [1,2]
π(forsome2) = [1,2]
π(and) = []
π(and1) = [1]
π(or) = []
π(or1) = [1]
π(forall) = []
π(forall1) = [1]
π(cons) = []
π(cons1) = [1]
π(forsome) = []
π(forsome1) = [1]

all of the following rules can be deleted.
and2(true,true) → true (21)
and2(x,false) → false (22)
and2(false,y) → false (23)
or2(true,y) → true (24)
or2(x,true) → true (25)
or2(false,false) → false (26)
forall2(p,nil) → true (27)
forall2(p,cons2(x,xs)) → and2(app(p,x),forall2(p,xs)) (28)
forsome2(p,nil) → false (29)
forsome2(p,cons2(x,xs)) → or2(app(p,x),forsome2(p,xs)) (30)
app(and,y1) → and1(y1) (11)
app(and1(x0),y1) → and2(x0,y1) (12)
app(or,y1) → or1(y1) (13)
app(or1(x0),y1) → or2(x0,y1) (14)
app(forall,y1) → forall1(y1) (15)
app(forall1(x0),y1) → forall2(x0,y1) (16)
app(cons,y1) → cons1(y1) (17)
app(cons1(x0),y1) → cons2(x0,y1) (18)
app(forsome,y1) → forsome1(y1) (19)
app(forsome1(x0),y1) → forsome2(x0,y1) (20)

1.1.1 R is empty

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