YES

The TRS could be proven terminating. The proof took 1671 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (463ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4 (335ms).
 |    | – Problem 4 was processed with processor DependencyGraph (19ms).
 |    |    | – Problem 5 was processed with processor PolynomialLinearRange4 (127ms).
 |    |    |    | – Problem 8 was processed with processor DependencyGraph (4ms).
 |    |    | – Problem 6 was processed with processor PolynomialLinearRange4 (13ms).
 |    |    | – Problem 7 was processed with processor PolynomialLinearRange4 (17ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4 (253ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

U11#(tt, V1, V2)U12#(isNat(V1), V2)isNat#(plus(V1, V2))U11#(and(isNatKind(V1), isNatKind(V2)), V1, V2)
plus#(N, 0)isNat#(N)U12#(tt, V2)isNat#(V2)
isNat#(s(V1))isNatKind#(V1)isNat#(s(V1))U21#(isNatKind(V1), V1)
plus#(N, s(M))and#(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N)))T(isNat(x_1))T(x_1)
T(and(x_1, x_2))T(x_1)plus#(N, s(M))isNat#(M)
U12#(tt, V2)U13#(isNat(V2))isNatKind#(s(V1))isNatKind#(V1)
isNatKind#(plus(V1, V2))and#(isNatKind(V1), isNatKind(V2))T(isNatKind(V2))isNatKind#(V2)
isNat#(plus(V1, V2))and#(isNatKind(V1), isNatKind(V2))U21#(tt, V1)isNat#(V1)
T(isNatKind(M))isNatKind#(M)and#(tt, X)T(X)
U31#(tt, N)T(N)isNatKind#(plus(V1, V2))isNatKind#(V1)
U21#(tt, V1)U22#(isNat(V1))T(and(isNat(N), isNatKind(N)))and#(isNat(N), isNatKind(N))
plus#(N, 0)and#(isNat(N), isNatKind(N))T(isNatKind(N))isNatKind#(N)
U41#(tt, M, N)plus#(N, M)plus#(N, s(M))and#(isNat(M), isNatKind(M))
U11#(tt, V1, V2)isNat#(V1)U41#(tt, M, N)T(M)
plus#(N, 0)U31#(and(isNat(N), isNatKind(N)), N)plus#(N, s(M))U41#(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)
isNat#(plus(V1, V2))isNatKind#(V1)U41#(tt, M, N)T(N)
T(isNat(N))isNat#(N)T(isNatKind(x_1))T(x_1)
T(and(x_1, x_2))T(x_2)

Rewrite Rules

U11(tt, V1, V2)U12(isNat(V1), V2)U12(tt, V2)U13(isNat(V2))
U13(tt)ttU21(tt, V1)U22(isNat(V1))
U22(tt)ttU31(tt, N)N
U41(tt, M, N)s(plus(N, M))and(tt, X)X
isNat(0)ttisNat(plus(V1, V2))U11(and(isNatKind(V1), isNatKind(V2)), V1, V2)
isNat(s(V1))U21(isNatKind(V1), V1)isNatKind(0)tt
isNatKind(plus(V1, V2))and(isNatKind(V1), isNatKind(V2))isNatKind(s(V1))isNatKind(V1)
plus(N, 0)U31(and(isNat(N), isNatKind(N)), N)plus(N, s(M))U41(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)

Original Signature

Termination of terms over the following signature is verified: plus, isNatKind, and, isNat, 0, s, tt, U41, U11, U12, U13, U31, U21, U22

Strategy

Context-sensitive strategy:
μ(isNat) = μ(T) = μ(isNatKind) = μ(0) = μ(isNatKind#) = μ(tt) = μ(isNat#) = ∅
μ(U11#) = μ(U13#) = μ(U31#) = μ(U21#) = μ(and#) = μ(U41#) = μ(U41) = μ(U21) = μ(U22) = μ(U12#) = μ(U22#) = μ(and) = μ(s) = μ(U11) = μ(U12) = μ(U31) = μ(U13) = {1}
μ(plus) = μ(plus#) = {1, 2}


The following SCCs where found

plus#(N, s(M)) → U41#(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)U41#(tt, M, N) → plus#(N, M)

U11#(tt, V1, V2) → U12#(isNat(V1), V2)and#(tt, X) → T(X)
T(isNatKind(M)) → isNatKind#(M)isNatKind#(plus(V1, V2)) → isNatKind#(V1)
isNat#(plus(V1, V2)) → U11#(and(isNatKind(V1), isNatKind(V2)), V1, V2)U12#(tt, V2) → isNat#(V2)
T(and(isNat(N), isNatKind(N))) → and#(isNat(N), isNatKind(N))isNat#(s(V1)) → isNatKind#(V1)
isNat#(s(V1)) → U21#(isNatKind(V1), V1)T(isNatKind(N)) → isNatKind#(N)
T(isNat(x_1)) → T(x_1)T(and(x_1, x_2)) → T(x_1)
U11#(tt, V1, V2) → isNat#(V1)isNatKind#(s(V1)) → isNatKind#(V1)
isNatKind#(plus(V1, V2)) → and#(isNatKind(V1), isNatKind(V2))isNat#(plus(V1, V2)) → isNatKind#(V1)
T(isNat(N)) → isNat#(N)T(isNatKind(V2)) → isNatKind#(V2)
isNat#(plus(V1, V2)) → and#(isNatKind(V1), isNatKind(V2))T(isNatKind(x_1)) → T(x_1)
T(and(x_1, x_2)) → T(x_2)U21#(tt, V1) → isNat#(V1)

Problem 2: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

U11#(tt, V1, V2)U12#(isNat(V1), V2)isNatKind#(plus(V1, V2))isNatKind#(V1)
T(isNatKind(M))isNatKind#(M)and#(tt, X)T(X)
isNat#(plus(V1, V2))U11#(and(isNatKind(V1), isNatKind(V2)), V1, V2)T(and(isNat(N), isNatKind(N)))and#(isNat(N), isNatKind(N))
U12#(tt, V2)isNat#(V2)isNat#(s(V1))isNatKind#(V1)
isNat#(s(V1))U21#(isNatKind(V1), V1)T(isNatKind(N))isNatKind#(N)
T(and(x_1, x_2))T(x_1)T(isNat(x_1))T(x_1)
U11#(tt, V1, V2)isNat#(V1)isNatKind#(s(V1))isNatKind#(V1)
isNatKind#(plus(V1, V2))and#(isNatKind(V1), isNatKind(V2))isNat#(plus(V1, V2))isNatKind#(V1)
T(isNat(N))isNat#(N)T(isNatKind(V2))isNatKind#(V2)
isNat#(plus(V1, V2))and#(isNatKind(V1), isNatKind(V2))T(isNatKind(x_1))T(x_1)
T(and(x_1, x_2))T(x_2)U21#(tt, V1)isNat#(V1)

Rewrite Rules

U11(tt, V1, V2)U12(isNat(V1), V2)U12(tt, V2)U13(isNat(V2))
U13(tt)ttU21(tt, V1)U22(isNat(V1))
U22(tt)ttU31(tt, N)N
U41(tt, M, N)s(plus(N, M))and(tt, X)X
isNat(0)ttisNat(plus(V1, V2))U11(and(isNatKind(V1), isNatKind(V2)), V1, V2)
isNat(s(V1))U21(isNatKind(V1), V1)isNatKind(0)tt
isNatKind(plus(V1, V2))and(isNatKind(V1), isNatKind(V2))isNatKind(s(V1))isNatKind(V1)
plus(N, 0)U31(and(isNat(N), isNatKind(N)), N)plus(N, s(M))U41(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)

Original Signature

Termination of terms over the following signature is verified: plus, isNatKind, and, isNat, 0, s, tt, U41, U11, U12, U13, U31, U21, U22

Strategy

Context-sensitive strategy:
μ(isNat) = μ(T) = μ(isNatKind) = μ(0) = μ(isNatKind#) = μ(tt) = μ(isNat#) = ∅
μ(U11#) = μ(U31#) = μ(U13#) = μ(U21#) = μ(and#) = μ(U41#) = μ(U41) = μ(U21) = μ(U22) = μ(U12#) = μ(U22#) = μ(and) = μ(s) = μ(U11) = μ(U12) = μ(U13) = μ(U31) = {1}
μ(plus) = μ(plus#) = {1, 2}


Polynomial Interpretation

Standard Usable rules

isNat(0)ttplus(N, s(M))U41(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)
U21(tt, V1)U22(isNat(V1))plus(N, 0)U31(and(isNat(N), isNatKind(N)), N)
U41(tt, M, N)s(plus(N, M))isNat(s(V1))U21(isNatKind(V1), V1)
isNatKind(s(V1))isNatKind(V1)U31(tt, N)N
U11(tt, V1, V2)U12(isNat(V1), V2)U13(tt)tt
isNat(plus(V1, V2))U11(and(isNatKind(V1), isNatKind(V2)), V1, V2)and(tt, X)X
isNatKind(plus(V1, V2))and(isNatKind(V1), isNatKind(V2))isNatKind(0)tt
U12(tt, V2)U13(isNat(V2))U22(tt)tt

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

isNatKind#(plus(V1, V2))isNatKind#(V1)T(isNatKind(M))isNatKind#(M)
isNat#(plus(V1, V2))U11#(and(isNatKind(V1), isNatKind(V2)), V1, V2)T(and(isNat(N), isNatKind(N)))and#(isNat(N), isNatKind(N))
isNat#(s(V1))isNatKind#(V1)T(isNatKind(N))isNatKind#(N)
T(isNat(x_1))T(x_1)isNat#(plus(V1, V2))isNatKind#(V1)
T(isNat(N))isNat#(N)T(isNatKind(V2))isNatKind#(V2)
isNat#(plus(V1, V2))and#(isNatKind(V1), isNatKind(V2))

Problem 4: DependencyGraph



Dependency Pair Problem

Dependency Pairs

isNatKind#(s(V1))isNatKind#(V1)U11#(tt, V1, V2)U12#(isNat(V1), V2)
and#(tt, X)T(X)isNatKind#(plus(V1, V2))and#(isNatKind(V1), isNatKind(V2))
U12#(tt, V2)isNat#(V2)isNat#(s(V1))U21#(isNatKind(V1), V1)
T(isNatKind(x_1))T(x_1)T(and(x_1, x_2))T(x_1)
T(and(x_1, x_2))T(x_2)U21#(tt, V1)isNat#(V1)
U11#(tt, V1, V2)isNat#(V1)

Rewrite Rules

U11(tt, V1, V2)U12(isNat(V1), V2)U12(tt, V2)U13(isNat(V2))
U13(tt)ttU21(tt, V1)U22(isNat(V1))
U22(tt)ttU31(tt, N)N
U41(tt, M, N)s(plus(N, M))and(tt, X)X
isNat(0)ttisNat(plus(V1, V2))U11(and(isNatKind(V1), isNatKind(V2)), V1, V2)
isNat(s(V1))U21(isNatKind(V1), V1)isNatKind(0)tt
isNatKind(plus(V1, V2))and(isNatKind(V1), isNatKind(V2))isNatKind(s(V1))isNatKind(V1)
plus(N, 0)U31(and(isNat(N), isNatKind(N)), N)plus(N, s(M))U41(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)

Original Signature

Termination of terms over the following signature is verified: plus, isNatKind, and, isNat, 0, s, tt, U41, U11, U12, U13, U31, U21, U22

Strategy

Context-sensitive strategy:
μ(isNat) = μ(T) = μ(isNatKind) = μ(0) = μ(isNatKind#) = μ(tt) = μ(isNat#) = ∅
μ(U11#) = μ(U13#) = μ(U31#) = μ(U21#) = μ(and#) = μ(U41#) = μ(U41) = μ(U21) = μ(U22) = μ(U12#) = μ(U22#) = μ(and) = μ(s) = μ(U11) = μ(U12) = μ(U31) = μ(U13) = {1}
μ(plus) = μ(plus#) = {1, 2}


The following SCCs where found

isNatKind#(s(V1)) → isNatKind#(V1)

T(and(x_1, x_2)) → T(x_1)T(isNatKind(x_1)) → T(x_1)
T(and(x_1, x_2)) → T(x_2)

isNat#(s(V1)) → U21#(isNatKind(V1), V1)U21#(tt, V1) → isNat#(V1)

Problem 5: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

isNat#(s(V1))U21#(isNatKind(V1), V1)U21#(tt, V1)isNat#(V1)

Rewrite Rules

U11(tt, V1, V2)U12(isNat(V1), V2)U12(tt, V2)U13(isNat(V2))
U13(tt)ttU21(tt, V1)U22(isNat(V1))
U22(tt)ttU31(tt, N)N
U41(tt, M, N)s(plus(N, M))and(tt, X)X
isNat(0)ttisNat(plus(V1, V2))U11(and(isNatKind(V1), isNatKind(V2)), V1, V2)
isNat(s(V1))U21(isNatKind(V1), V1)isNatKind(0)tt
isNatKind(plus(V1, V2))and(isNatKind(V1), isNatKind(V2))isNatKind(s(V1))isNatKind(V1)
plus(N, 0)U31(and(isNat(N), isNatKind(N)), N)plus(N, s(M))U41(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)

Original Signature

Termination of terms over the following signature is verified: plus, isNatKind, and, isNat, 0, s, tt, U41, U11, U12, U13, U31, U21, U22

Strategy

Context-sensitive strategy:
μ(isNat) = μ(T) = μ(isNatKind) = μ(0) = μ(isNatKind#) = μ(tt) = μ(isNat#) = ∅
μ(U11#) = μ(U31#) = μ(U13#) = μ(U21#) = μ(and#) = μ(U41#) = μ(U41) = μ(U21) = μ(U22) = μ(U12#) = μ(U22#) = μ(and) = μ(s) = μ(U11) = μ(U12) = μ(U13) = μ(U31) = {1}
μ(plus) = μ(plus#) = {1, 2}


Polynomial Interpretation

Standard Usable rules

isNat(0)ttplus(N, s(M))U41(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)
U21(tt, V1)U22(isNat(V1))plus(N, 0)U31(and(isNat(N), isNatKind(N)), N)
U41(tt, M, N)s(plus(N, M))isNat(s(V1))U21(isNatKind(V1), V1)
isNatKind(s(V1))isNatKind(V1)U31(tt, N)N
U11(tt, V1, V2)U12(isNat(V1), V2)U13(tt)tt
isNat(plus(V1, V2))U11(and(isNatKind(V1), isNatKind(V2)), V1, V2)and(tt, X)X
isNatKind(plus(V1, V2))and(isNatKind(V1), isNatKind(V2))isNatKind(0)tt
U12(tt, V2)U13(isNat(V2))U22(tt)tt

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

isNat#(s(V1))U21#(isNatKind(V1), V1)

Problem 8: DependencyGraph



Dependency Pair Problem

Dependency Pairs

U21#(tt, V1)isNat#(V1)

Rewrite Rules

U11(tt, V1, V2)U12(isNat(V1), V2)U12(tt, V2)U13(isNat(V2))
U13(tt)ttU21(tt, V1)U22(isNat(V1))
U22(tt)ttU31(tt, N)N
U41(tt, M, N)s(plus(N, M))and(tt, X)X
isNat(0)ttisNat(plus(V1, V2))U11(and(isNatKind(V1), isNatKind(V2)), V1, V2)
isNat(s(V1))U21(isNatKind(V1), V1)isNatKind(0)tt
isNatKind(plus(V1, V2))and(isNatKind(V1), isNatKind(V2))isNatKind(s(V1))isNatKind(V1)
plus(N, 0)U31(and(isNat(N), isNatKind(N)), N)plus(N, s(M))U41(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)

Original Signature

Termination of terms over the following signature is verified: plus, isNatKind, and, isNat, 0, s, tt, U41, U11, U12, U13, U31, U21, U22

Strategy

Context-sensitive strategy:
μ(isNat) = μ(T) = μ(isNatKind) = μ(0) = μ(isNatKind#) = μ(tt) = μ(isNat#) = ∅
μ(U11#) = μ(U13#) = μ(U31#) = μ(U21#) = μ(and#) = μ(U41#) = μ(U41) = μ(U21) = μ(U22) = μ(U12#) = μ(U22#) = μ(and) = μ(s) = μ(U11) = μ(U12) = μ(U31) = μ(U13) = {1}
μ(plus) = μ(plus#) = {1, 2}


There are no SCCs!

Problem 6: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

isNatKind#(s(V1))isNatKind#(V1)

Rewrite Rules

U11(tt, V1, V2)U12(isNat(V1), V2)U12(tt, V2)U13(isNat(V2))
U13(tt)ttU21(tt, V1)U22(isNat(V1))
U22(tt)ttU31(tt, N)N
U41(tt, M, N)s(plus(N, M))and(tt, X)X
isNat(0)ttisNat(plus(V1, V2))U11(and(isNatKind(V1), isNatKind(V2)), V1, V2)
isNat(s(V1))U21(isNatKind(V1), V1)isNatKind(0)tt
isNatKind(plus(V1, V2))and(isNatKind(V1), isNatKind(V2))isNatKind(s(V1))isNatKind(V1)
plus(N, 0)U31(and(isNat(N), isNatKind(N)), N)plus(N, s(M))U41(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)

Original Signature

Termination of terms over the following signature is verified: plus, isNatKind, and, isNat, 0, s, tt, U41, U11, U12, U13, U31, U21, U22

Strategy

Context-sensitive strategy:
μ(isNat) = μ(T) = μ(isNatKind) = μ(0) = μ(isNatKind#) = μ(tt) = μ(isNat#) = ∅
μ(U11#) = μ(U31#) = μ(U13#) = μ(U21#) = μ(and#) = μ(U41#) = μ(U41) = μ(U21) = μ(U22) = μ(U12#) = μ(U22#) = μ(and) = μ(s) = μ(U11) = μ(U12) = μ(U13) = μ(U31) = {1}
μ(plus) = μ(plus#) = {1, 2}


Polynomial Interpretation

There are no usable rules

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

isNatKind#(s(V1))isNatKind#(V1)

Problem 7: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(and(x_1, x_2))T(x_1)T(isNatKind(x_1))T(x_1)
T(and(x_1, x_2))T(x_2)

Rewrite Rules

U11(tt, V1, V2)U12(isNat(V1), V2)U12(tt, V2)U13(isNat(V2))
U13(tt)ttU21(tt, V1)U22(isNat(V1))
U22(tt)ttU31(tt, N)N
U41(tt, M, N)s(plus(N, M))and(tt, X)X
isNat(0)ttisNat(plus(V1, V2))U11(and(isNatKind(V1), isNatKind(V2)), V1, V2)
isNat(s(V1))U21(isNatKind(V1), V1)isNatKind(0)tt
isNatKind(plus(V1, V2))and(isNatKind(V1), isNatKind(V2))isNatKind(s(V1))isNatKind(V1)
plus(N, 0)U31(and(isNat(N), isNatKind(N)), N)plus(N, s(M))U41(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)

Original Signature

Termination of terms over the following signature is verified: plus, isNatKind, and, isNat, 0, s, tt, U41, U11, U12, U13, U31, U21, U22

Strategy

Context-sensitive strategy:
μ(isNat) = μ(T) = μ(isNatKind) = μ(0) = μ(isNatKind#) = μ(tt) = μ(isNat#) = ∅
μ(U11#) = μ(U31#) = μ(U13#) = μ(U21#) = μ(and#) = μ(U41#) = μ(U41) = μ(U21) = μ(U22) = μ(U12#) = μ(U22#) = μ(and) = μ(s) = μ(U11) = μ(U12) = μ(U13) = μ(U31) = {1}
μ(plus) = μ(plus#) = {1, 2}


Polynomial Interpretation

There are no usable rules

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

T(isNatKind(x_1))T(x_1)T(and(x_1, x_2))T(x_1)
T(and(x_1, x_2))T(x_2)

Problem 3: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

plus#(N, s(M))U41#(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)U41#(tt, M, N)plus#(N, M)

Rewrite Rules

U11(tt, V1, V2)U12(isNat(V1), V2)U12(tt, V2)U13(isNat(V2))
U13(tt)ttU21(tt, V1)U22(isNat(V1))
U22(tt)ttU31(tt, N)N
U41(tt, M, N)s(plus(N, M))and(tt, X)X
isNat(0)ttisNat(plus(V1, V2))U11(and(isNatKind(V1), isNatKind(V2)), V1, V2)
isNat(s(V1))U21(isNatKind(V1), V1)isNatKind(0)tt
isNatKind(plus(V1, V2))and(isNatKind(V1), isNatKind(V2))isNatKind(s(V1))isNatKind(V1)
plus(N, 0)U31(and(isNat(N), isNatKind(N)), N)plus(N, s(M))U41(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)

Original Signature

Termination of terms over the following signature is verified: plus, isNatKind, and, isNat, 0, s, tt, U41, U11, U12, U13, U31, U21, U22

Strategy

Context-sensitive strategy:
μ(isNat) = μ(T) = μ(isNatKind) = μ(0) = μ(isNatKind#) = μ(tt) = μ(isNat#) = ∅
μ(U11#) = μ(U31#) = μ(U13#) = μ(U21#) = μ(and#) = μ(U41#) = μ(U41) = μ(U21) = μ(U22) = μ(U12#) = μ(U22#) = μ(and) = μ(s) = μ(U11) = μ(U12) = μ(U13) = μ(U31) = {1}
μ(plus) = μ(plus#) = {1, 2}


Polynomial Interpretation

Standard Usable rules

isNat(0)ttplus(N, s(M))U41(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)
U21(tt, V1)U22(isNat(V1))plus(N, 0)U31(and(isNat(N), isNatKind(N)), N)
U41(tt, M, N)s(plus(N, M))isNat(s(V1))U21(isNatKind(V1), V1)
isNatKind(s(V1))isNatKind(V1)U31(tt, N)N
U11(tt, V1, V2)U12(isNat(V1), V2)U13(tt)tt
isNat(plus(V1, V2))U11(and(isNatKind(V1), isNatKind(V2)), V1, V2)and(tt, X)X
isNatKind(plus(V1, V2))and(isNatKind(V1), isNatKind(V2))isNatKind(0)tt
U12(tt, V2)U13(isNat(V2))U22(tt)tt

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

plus#(N, s(M))U41#(and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N)U41#(tt, M, N)plus#(N, M)