Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/MYNAT_nosorts_noand_GM)

The rewrite relation of the following TRS is considered.

a__U11(tt,M,N) a__U12(tt,M,N) (1)
a__U12(tt,M,N) s(a__plus(mark(N),mark(M))) (2)
a__U21(tt,M,N) a__U22(tt,M,N) (3)
a__U22(tt,M,N) a__plus(a__x(mark(N),mark(M)),mark(N)) (4)
a__plus(N,0) mark(N) (5)
a__plus(N,s(M)) a__U11(tt,M,N) (6)
a__x(N,0) 0 (7)
a__x(N,s(M)) a__U21(tt,M,N) (8)
mark(U11(X1,X2,X3)) a__U11(mark(X1),X2,X3) (9)
mark(U12(X1,X2,X3)) a__U12(mark(X1),X2,X3) (10)
mark(plus(X1,X2)) a__plus(mark(X1),mark(X2)) (11)
mark(U21(X1,X2,X3)) a__U21(mark(X1),X2,X3) (12)
mark(U22(X1,X2,X3)) a__U22(mark(X1),X2,X3) (13)
mark(x(X1,X2)) a__x(mark(X1),mark(X2)) (14)
mark(tt) tt (15)
mark(s(X)) s(mark(X)) (16)
mark(0) 0 (17)
a__U11(X1,X2,X3) U11(X1,X2,X3) (18)
a__U12(X1,X2,X3) U12(X1,X2,X3) (19)
a__plus(X1,X2) plus(X1,X2) (20)
a__U21(X1,X2,X3) U21(X1,X2,X3) (21)
a__U22(X1,X2,X3) U22(X1,X2,X3) (22)
a__x(X1,X2) x(X1,X2) (23)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the
prec(a__U11) = 1 stat(a__U11) = mul
prec(tt) = 0 stat(tt) = mul
prec(a__U12) = 1 stat(a__U12) = mul
prec(s) = 0 stat(s) = mul
prec(a__plus) = 1 stat(a__plus) = mul
prec(a__U21) = 2 stat(a__U21) = lex
prec(a__U22) = 2 stat(a__U22) = lex
prec(a__x) = 2 stat(a__x) = lex
prec(0) = 3 stat(0) = mul
prec(U11) = 1 stat(U11) = mul
prec(U12) = 1 stat(U12) = mul
prec(plus) = 1 stat(plus) = mul
prec(U21) = 2 stat(U21) = lex
prec(U22) = 2 stat(U22) = lex
prec(x) = 2 stat(x) = lex

π(a__U11) = [1,2,3]
π(tt) = []
π(a__U12) = [1,2,3]
π(s) = [1]
π(a__plus) = [1,2]
π(mark) = 1
π(a__U21) = [3,2,1]
π(a__U22) = [3,2,1]
π(a__x) = [1,2]
π(0) = []
π(U11) = [1,2,3]
π(U12) = [1,2,3]
π(plus) = [1,2]
π(U21) = [3,2,1]
π(U22) = [3,2,1]
π(x) = [1,2]

all of the following rules can be deleted.
a__U12(tt,M,N) s(a__plus(mark(N),mark(M))) (2)
a__U22(tt,M,N) a__plus(a__x(mark(N),mark(M)),mark(N)) (4)
a__plus(N,0) mark(N) (5)
a__plus(N,s(M)) a__U11(tt,M,N) (6)
a__x(N,0) 0 (7)
a__x(N,s(M)) a__U21(tt,M,N) (8)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(tt) = 1 weight(tt) = 1
prec(0) = 14 weight(0) = 1
prec(mark) = 15 weight(mark) = 0
prec(s) = 7 weight(s) = 1
prec(a__U11) = 11 weight(a__U11) = 0
prec(a__U12) = 10 weight(a__U12) = 0
prec(a__U21) = 3 weight(a__U21) = 1
prec(a__U22) = 13 weight(a__U22) = 0
prec(U11) = 0 weight(U11) = 0
prec(U12) = 9 weight(U12) = 0
prec(plus) = 4 weight(plus) = 0
prec(a__plus) = 5 weight(a__plus) = 0
prec(U21) = 2 weight(U21) = 1
prec(U22) = 12 weight(U22) = 0
prec(x) = 6 weight(x) = 0
prec(a__x) = 8 weight(a__x) = 0
all of the following rules can be deleted.
a__U11(tt,M,N) a__U12(tt,M,N) (1)
a__U21(tt,M,N) a__U22(tt,M,N) (3)
mark(U11(X1,X2,X3)) a__U11(mark(X1),X2,X3) (9)
mark(U12(X1,X2,X3)) a__U12(mark(X1),X2,X3) (10)
mark(plus(X1,X2)) a__plus(mark(X1),mark(X2)) (11)
mark(U21(X1,X2,X3)) a__U21(mark(X1),X2,X3) (12)
mark(U22(X1,X2,X3)) a__U22(mark(X1),X2,X3) (13)
mark(x(X1,X2)) a__x(mark(X1),mark(X2)) (14)
mark(tt) tt (15)
mark(s(X)) s(mark(X)) (16)
mark(0) 0 (17)
a__U11(X1,X2,X3) U11(X1,X2,X3) (18)
a__U12(X1,X2,X3) U12(X1,X2,X3) (19)
a__plus(X1,X2) plus(X1,X2) (20)
a__U21(X1,X2,X3) U21(X1,X2,X3) (21)
a__U22(X1,X2,X3) U22(X1,X2,X3) (22)
a__x(X1,X2) x(X1,X2) (23)

1.1.1 R is empty

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