YES Termination w.r.t. Q proof of Transformed_CSR_04_PEANO_complete_FR.ari

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 QTRSRRRProof (⇔, 131 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 RisEmptyProof (⇔, 0 ms)
↳6 YES

(0) Obligation:

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

U11(tt, V1, V2) → U12(isNat(activate(V1)), activate(V2))
U12(tt, V2) → U13(isNat(activate(V2)))
U13(tt) → tt
U21(tt, V1) → U22(isNat(activate(V1)))
U22(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(and(isNatKind(activate(V1)), n__isNatKind(activate(V2))), activate(V1), activate(V2))
isNat(n__s(V1)) → U21(isNatKind(activate(V1)), activate(V1))
isNatKind(n__0) → tt
isNatKind(n__plus(V1, V2)) → and(isNatKind(activate(V1)), n__isNatKind(activate(V2)))
isNatKind(n__s(V1)) → isNatKind(activate(V1))
plus(N, 0) → U31(and(isNat(N), n__isNatKind(N)), N)
plus(N, s(M)) → U41(and(and(isNat(M), n__isNatKind(M)), n__and(n__isNat(N), n__isNatKind(N))), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNatKind(X) → n__isNatKind(X)
s(X) → n__s(X)
and(X1, X2) → n__and(X1, X2)
isNat(X) → n__isNat(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(activate(X1), activate(X2))
activate(n__isNatKind(X)) → isNatKind(X)
activate(n__s(X)) → s(activate(X))
activate(n__and(X1, X2)) → and(activate(X1), X2)
activate(n__isNat(X)) → isNat(X)
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
U11(x1, x2, x3)  =  U11(x1, x2, x3)
tt  =  tt
U12(x1, x2)  =  U12(x1, x2)
isNat(x1)  =  x1
activate(x1)  =  x1
U13(x1)  =  x1
U21(x1, x2)  =  U21(x1, x2)
U22(x1)  =  x1
U31(x1, x2)  =  U31(x1, x2)
U41(x1, x2, x3)  =  U41(x1, x2, x3)
s(x1)  =  s(x1)
plus(x1, x2)  =  plus(x1, x2)
and(x1, x2)  =  and(x1, x2)
n__0  =  n__0
n__plus(x1, x2)  =  n__plus(x1, x2)
isNatKind(x1)  =  x1
n__isNatKind(x1)  =  x1
n__s(x1)  =  n__s(x1)
0  =  0
n__and(x1, x2)  =  n__and(x1, x2)
n__isNat(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
[U413, plus2, nplus2] > U113 > U122 > [U212, U312]
[U413, plus2, nplus2] > [s1, ns1] > [and2, nand2] > [U212, U312]
[n0, 0] > tt > U122 > [U212, U312]
[n0, 0] > [and2, nand2] > [U212, U312]

Status:
U113: [1,2,3]
tt: multiset
U122: [2,1]
U212: [1,2]
U312: multiset
U413: [2,3,1]
s1: [1]
plus2: [2,1]
and2: multiset
n0: multiset
nplus2: [2,1]
ns1: [1]
0: multiset
nand2: multiset

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

U11(tt, V1, V2) → U12(isNat(activate(V1)), activate(V2))
U12(tt, V2) → U13(isNat(activate(V2)))
U21(tt, V1) → U22(isNat(activate(V1)))
U31(tt, N) → activate(N)
U41(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(and(isNatKind(activate(V1)), n__isNatKind(activate(V2))), activate(V1), activate(V2))
isNat(n__s(V1)) → U21(isNatKind(activate(V1)), activate(V1))
isNatKind(n__0) → tt
isNatKind(n__plus(V1, V2)) → and(isNatKind(activate(V1)), n__isNatKind(activate(V2)))
isNatKind(n__s(V1)) → isNatKind(activate(V1))
plus(N, 0) → U31(and(isNat(N), n__isNatKind(N)), N)
plus(N, s(M)) → U41(and(and(isNat(M), n__isNatKind(M)), n__and(n__isNat(N), n__isNatKind(N))), M, N)


(2) Obligation:

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

U13(tt) → tt
U22(tt) → tt
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNatKind(X) → n__isNatKind(X)
s(X) → n__s(X)
and(X1, X2) → n__and(X1, X2)
isNat(X) → n__isNat(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(activate(X1), activate(X2))
activate(n__isNatKind(X)) → isNatKind(X)
activate(n__s(X)) → s(activate(X))
activate(n__and(X1, X2)) → and(activate(X1), X2)
activate(n__isNat(X)) → isNat(X)
activate(X) → X

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Knuth-Bendix order [KBO] with precedence:
activate1 > 0 > n0 > tt > plus2 > U221 > isNat1 > nisNat1 > and2 > nand2 > s1 > ns1 > U131 > isNatKind1 > nisNatKind1 > nplus2

and weight map:

tt=1
0=1
n__0=1
U13_1=1
U22_1=1
isNatKind_1=1
n__isNatKind_1=1
s_1=1
n__s_1=1
isNat_1=1
n__isNat_1=1
activate_1=0
plus_2=0
n__plus_2=0
and_2=0
n__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:

U13(tt) → tt
U22(tt) → tt
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNatKind(X) → n__isNatKind(X)
s(X) → n__s(X)
and(X1, X2) → n__and(X1, X2)
isNat(X) → n__isNat(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(activate(X1), activate(X2))
activate(n__isNatKind(X)) → isNatKind(X)
activate(n__s(X)) → s(activate(X))
activate(n__and(X1, X2)) → and(activate(X1), X2)
activate(n__isNat(X)) → isNat(X)
activate(X) → 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