The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

U11(tt, N) → N
U21(tt, M, N) → s(plus(N, M))
and(tt, X) → X
isNat(0) → tt
isNat(plus(V1, V2)) → and(isNat(V1), isNat(V2))
isNat(s(V1)) → isNat(V1)
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), isNat(N)), M, N)

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0) → z0
U21(tt, z0, z1) → s(plus(z1, z0))
and(tt, z0) → z0
isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
plus(z0, 0) → U11(isNat(z0), z0)
plus(z0, s(z1)) → U21(and(isNat(z1), isNat(z0)), z1, z0)

Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(plus(z0, z1)) → c4(AND(isNat(z0), isNat(z1)), ISNAT(z0), ISNAT(z1))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c6(U11'(isNat(z0), z0), ISNAT(z0))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), AND(isNat(z1), isNat(z0)), ISNAT(z1), ISNAT(z0))

S tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(plus(z0, z1)) → c4(AND(isNat(z0), isNat(z1)), ISNAT(z0), ISNAT(z1))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c6(U11'(isNat(z0), z0), ISNAT(z0))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), AND(isNat(z1), isNat(z0)), ISNAT(z1), ISNAT(z0))

K tuples:none
Defined Rule Symbols:

U11, U21, and, isNat, plus

Defined Pair Symbols:

U21', ISNAT, PLUS

Compound Symbols:

c1, c4, c5, c6, c7

(3) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)

Split RHS of tuples not part of any SCC

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0) → z0
U21(tt, z0, z1) → s(plus(z1, z0))
and(tt, z0) → z0
isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
plus(z0, 0) → U11(isNat(z0), z0)
plus(z0, s(z1)) → U21(and(isNat(z1), isNat(z0)), z1, z0)

Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(plus(z0, z1)) → c4(AND(isNat(z0), isNat(z1)), ISNAT(z0), ISNAT(z1))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), AND(isNat(z1), isNat(z0)), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c(U11'(isNat(z0), z0))
PLUS(z0, 0) → c(ISNAT(z0))

S tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(plus(z0, z1)) → c4(AND(isNat(z0), isNat(z1)), ISNAT(z0), ISNAT(z1))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), AND(isNat(z1), isNat(z0)), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c(U11'(isNat(z0), z0))
PLUS(z0, 0) → c(ISNAT(z0))

K tuples:none
Defined Rule Symbols:

U11, U21, and, isNat, plus

Defined Pair Symbols:

U21', ISNAT, PLUS

Compound Symbols:

c1, c4, c5, c7, c

(5) CdtGraphRemoveTrailingProof (BOTH BOUNDS(ID, ID) transformation)

Removed 3 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0) → z0
U21(tt, z0, z1) → s(plus(z1, z0))
and(tt, z0) → z0
isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
plus(z0, 0) → U11(isNat(z0), z0)
plus(z0, s(z1)) → U21(and(isNat(z1), isNat(z0)), z1, z0)

Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

S tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

K tuples:none
Defined Rule Symbols:

U11, U21, and, isNat, plus

Defined Pair Symbols:

U21', ISNAT, PLUS

Compound Symbols:

c1, c5, c, c4, c7, c

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

PLUS(z0, 0) → c(ISNAT(z0))

We considered the (Usable) Rules:

isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
and(tt, z0) → z0

And the Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ISNAT(x₁)) = 0   
POL(PLUS(x₁, x₂)) = [4]   
POL(U21'(x₁, x₂, x₃)) = [4]   
POL(and(x₁, x₂)) = [5] + [5]x₁ + x₂   
POL(c) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(isNat(x₁)) = [3]   
POL(plus(x₁, x₂)) = [4] + [5]x₁ + [2]x₂   
POL(s(x₁)) = 0   
POL(tt) = [2]   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0) → z0
U21(tt, z0, z1) → s(plus(z1, z0))
and(tt, z0) → z0
isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
plus(z0, 0) → U11(isNat(z0), z0)
plus(z0, s(z1)) → U21(and(isNat(z1), isNat(z0)), z1, z0)

Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

S tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

K tuples:

PLUS(z0, 0) → c(ISNAT(z0))

Defined Rule Symbols:

U11, U21, and, isNat, plus

Defined Pair Symbols:

U21', ISNAT, PLUS

Compound Symbols:

c1, c5, c, c4, c7, c

(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

PLUS(z0, 0) → c

We considered the (Usable) Rules:

isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
and(tt, z0) → z0

And the Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [2]   
POL(ISNAT(x₁)) = 0   
POL(PLUS(x₁, x₂)) = [1]   
POL(U21'(x₁, x₂, x₃)) = [1]   
POL(and(x₁, x₂)) = [3] + [5]x₁   
POL(c) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(isNat(x₁)) = [3]   
POL(plus(x₁, x₂)) = [3] + [2]x₁ + [2]x₂   
POL(s(x₁)) = 0   
POL(tt) = [3]   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0) → z0
U21(tt, z0, z1) → s(plus(z1, z0))
and(tt, z0) → z0
isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
plus(z0, 0) → U11(isNat(z0), z0)
plus(z0, s(z1)) → U21(and(isNat(z1), isNat(z0)), z1, z0)

Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

S tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))

K tuples:

PLUS(z0, 0) → c(ISNAT(z0))
PLUS(z0, 0) → c

Defined Rule Symbols:

U11, U21, and, isNat, plus

Defined Pair Symbols:

U21', ISNAT, PLUS

Compound Symbols:

c1, c5, c, c4, c7, c

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))

We considered the (Usable) Rules:

isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
and(tt, z0) → z0

And the Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [4]   
POL(ISNAT(x₁)) = 0   
POL(PLUS(x₁, x₂)) = [4]x₂   
POL(U21'(x₁, x₂, x₃)) = [4]x₂   
POL(and(x₁, x₂)) = 0   
POL(c) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(isNat(x₁)) = [3]x₁   
POL(plus(x₁, x₂)) = [2]x₁   
POL(s(x₁)) = [2] + x₁   
POL(tt) = [3]   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0) → z0
U21(tt, z0, z1) → s(plus(z1, z0))
and(tt, z0) → z0
isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
plus(z0, 0) → U11(isNat(z0), z0)
plus(z0, s(z1)) → U21(and(isNat(z1), isNat(z0)), z1, z0)

Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

S tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))

K tuples:

PLUS(z0, 0) → c(ISNAT(z0))
PLUS(z0, 0) → c
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))

Defined Rule Symbols:

U11, U21, and, isNat, plus

Defined Pair Symbols:

U21', ISNAT, PLUS

Compound Symbols:

c1, c5, c, c4, c7, c

(13) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
PLUS(z0, 0) → c(ISNAT(z0))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0) → z0
U21(tt, z0, z1) → s(plus(z1, z0))
and(tt, z0) → z0
isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
plus(z0, 0) → U11(isNat(z0), z0)
plus(z0, s(z1)) → U21(and(isNat(z1), isNat(z0)), z1, z0)

Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

S tuples:

ISNAT(s(z0)) → c5(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))

K tuples:

PLUS(z0, 0) → c(ISNAT(z0))
PLUS(z0, 0) → c
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
U21'(tt, z0, z1) → c1(PLUS(z1, z0))

Defined Rule Symbols:

U11, U21, and, isNat, plus

Defined Pair Symbols:

U21', ISNAT, PLUS

Compound Symbols:

c1, c5, c, c4, c7, c

(15) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ISNAT(s(z0)) → c5(ISNAT(z0))

We considered the (Usable) Rules:

isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
and(tt, z0) → z0

And the Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [2]   
POL(ISNAT(x₁)) = [2] + [2]x₁   
POL(PLUS(x₁, x₂)) = [2]x₂² + x₁·x₂   
POL(U21'(x₁, x₂, x₃)) = x₂·x₃ + [2]x₂²   
POL(and(x₁, x₂)) = 0   
POL(c) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(isNat(x₁)) = 0   
POL(plus(x₁, x₂)) = [1] + [2]x₁ + [2]x₂   
POL(s(x₁)) = [2] + x₁   
POL(tt) = 0   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0) → z0
U21(tt, z0, z1) → s(plus(z1, z0))
and(tt, z0) → z0
isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
plus(z0, 0) → U11(isNat(z0), z0)
plus(z0, s(z1)) → U21(and(isNat(z1), isNat(z0)), z1, z0)

Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

S tuples:

ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))

K tuples:

PLUS(z0, 0) → c(ISNAT(z0))
PLUS(z0, 0) → c
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))

Defined Rule Symbols:

U11, U21, and, isNat, plus

Defined Pair Symbols:

U21', ISNAT, PLUS

Compound Symbols:

c1, c5, c, c4, c7, c

(17) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))

We considered the (Usable) Rules:

isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
and(tt, z0) → z0

And the Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ISNAT(x₁)) = [2] + [2]x₁   
POL(PLUS(x₁, x₂)) = [3] + [2]x₁ + [2]x₂² + x₁·x₂   
POL(U21'(x₁, x₂, x₃)) = [3] + [3]x₂ + [2]x₃ + x₂·x₃ + [2]x₂²   
POL(and(x₁, x₂)) = 0   
POL(c) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(isNat(x₁)) = 0   
POL(plus(x₁, x₂)) = [2] + x₁ + [2]x₂   
POL(s(x₁)) = [3] + x₁   
POL(tt) = 0   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0) → z0
U21(tt, z0, z1) → s(plus(z1, z0))
and(tt, z0) → z0
isNat(0) → tt
isNat(plus(z0, z1)) → and(isNat(z0), isNat(z1))
isNat(s(z0)) → isNat(z0)
plus(z0, 0) → U11(isNat(z0), z0)
plus(z0, s(z1)) → U21(and(isNat(z1), isNat(z0)), z1, z0)

Tuples:

U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
PLUS(z0, 0) → c(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
PLUS(z0, 0) → c

S tuples:none
K tuples:

PLUS(z0, 0) → c(ISNAT(z0))
PLUS(z0, 0) → c
PLUS(z0, s(z1)) → c7(U21'(and(isNat(z1), isNat(z0)), z1, z0), ISNAT(z1), ISNAT(z0))
U21'(tt, z0, z1) → c1(PLUS(z1, z0))
ISNAT(s(z0)) → c5(ISNAT(z0))
ISNAT(plus(z0, z1)) → c4(ISNAT(z0), ISNAT(z1))

Defined Rule Symbols:

U11, U21, and, isNat, plus

Defined Pair Symbols:

U21', ISNAT, PLUS

Compound Symbols:

c1, c5, c, c4, c7, c

(19) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(20) BOUNDS(O(1), O(1))