YES Termination w.r.t. Q proof of Transformed_CSR_04_PEANO_nokinds_GM.ari

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
a__U11(x1, x2)  =  a__U11(x1, x2)
tt  =  tt
mark(x1)  =  x1
a__U21(x1, x2, x3)  =  a__U21(x1, x2, x3)
s(x1)  =  s(x1)
a__plus(x1, x2)  =  a__plus(x1, x2)
a__and(x1, x2)  =  a__and(x1, x2)
a__isNat(x1)  =  a__isNat(x1)
0  =  0
plus(x1, x2)  =  plus(x1, x2)
isNat(x1)  =  isNat(x1)
U11(x1, x2)  =  U11(x1, x2)
U21(x1, x2, x3)  =  U21(x1, x2, x3)
and(x1, x2)  =  and(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:
[tt, aU213, aplus2, 0, plus2, U213] > [aU112, U112]
[tt, aU213, aplus2, 0, plus2, U213] > [s1, aand2, aisNat1, isNat1, and2]

Status:
aU112: multiset
tt: multiset
aU213: [2,3,1]
s1: [1]
aplus2: [2,1]
aand2: [1,2]
aisNat1: multiset
0: multiset
plus2: [2,1]
isNat1: multiset
U112: multiset
U213: [2,3,1]
and2: [1,2]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)


(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Knuth-Bendix order [KBO] with precedence:
mark1 > 0 > s1 > aand2 > and2 > tt > aU213 > U213 > aisNat1 > aU112 > isNat1 > aplus2 > U112 > plus2

and weight map:

tt=1
0=1
mark_1=0
isNat_1=1
a__isNat_1=1
s_1=1
U11_2=0
a__U11_2=0
U21_3=0
a__U21_3=0
plus_2=0
a__plus_2=0
and_2=0
a__and_2=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)


(4) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(5) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(6) YES