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

(0) Obligation:

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

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
0 → n__0
s(X) → n__s(X)
diff(X1, X2) → n__diff(X1, X2)
p(X) → n__p(X)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__diff(X1, X2)) → diff(activate(X1), activate(X2))
activate(n__p(X)) → p(activate(X))
activate(X) → X

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(0) → 0
p(s(z0)) → z0
p(z0) → n__p(z0)
leq(0, z0) → true
leq(s(z0), 0) → false
leq(s(z0), s(z1)) → leq(z0, z1)
if(true, z0, z1) → activate(z0)
if(false, z0, z1) → activate(z1)
diff(z0, z1) → if(leq(z0, z1), n__0, n__s(n__diff(n__p(z0), z1)))
diff(z0, z1) → n__diff(z0, z1)
0 → n__0
s(z0) → n__s(z0)
activate(n__0) → 0
activate(n__s(z0)) → s(activate(z0))
activate(n__diff(z0, z1)) → diff(activate(z0), activate(z1))
activate(n__p(z0)) → p(activate(z0))
activate(z0) → z0

Tuples:

P(0) → c(0')
LEQ(s(z0), s(z1)) → c5(LEQ(z0, z1))
IF(true, z0, z1) → c6(ACTIVATE(z0))
IF(false, z0, z1) → c7(ACTIVATE(z1))
DIFF(z0, z1) → c8(IF(leq(z0, z1), n__0, n__s(n__diff(n__p(z0), z1))), LEQ(z0, z1))
ACTIVATE(n__0) → c12(0')
ACTIVATE(n__s(z0)) → c13(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__diff(z0, z1)) → c14(DIFF(activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))
ACTIVATE(n__p(z0)) → c15(P(activate(z0)), ACTIVATE(z0))

S tuples:

P(0) → c(0')
LEQ(s(z0), s(z1)) → c5(LEQ(z0, z1))
IF(true, z0, z1) → c6(ACTIVATE(z0))
IF(false, z0, z1) → c7(ACTIVATE(z1))
DIFF(z0, z1) → c8(IF(leq(z0, z1), n__0, n__s(n__diff(n__p(z0), z1))), LEQ(z0, z1))
ACTIVATE(n__0) → c12(0')
ACTIVATE(n__s(z0)) → c13(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__diff(z0, z1)) → c14(DIFF(activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))
ACTIVATE(n__p(z0)) → c15(P(activate(z0)), ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

p, leq, if, diff, 0, s, activate

Defined Pair Symbols:

P, LEQ, IF, DIFF, ACTIVATE

Compound Symbols:

c, c5, c6, c7, c8, c12, c13, c14, c15

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 2 leading nodes:

IF(true, z0, z1) → c6(ACTIVATE(z0))
IF(false, z0, z1) → c7(ACTIVATE(z1))

Removed 4 trailing nodes:

P(0) → c(0')
ACTIVATE(n__0) → c12(0')
LEQ(s(z0), s(z1)) → c5(LEQ(z0, z1))
DIFF(z0, z1) → c8(IF(leq(z0, z1), n__0, n__s(n__diff(n__p(z0), z1))), LEQ(z0, z1))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(0) → 0
p(s(z0)) → z0
p(z0) → n__p(z0)
leq(0, z0) → true
leq(s(z0), 0) → false
leq(s(z0), s(z1)) → leq(z0, z1)
if(true, z0, z1) → activate(z0)
if(false, z0, z1) → activate(z1)
diff(z0, z1) → if(leq(z0, z1), n__0, n__s(n__diff(n__p(z0), z1)))
diff(z0, z1) → n__diff(z0, z1)
0 → n__0
s(z0) → n__s(z0)
activate(n__0) → 0
activate(n__s(z0)) → s(activate(z0))
activate(n__diff(z0, z1)) → diff(activate(z0), activate(z1))
activate(n__p(z0)) → p(activate(z0))
activate(z0) → z0

Tuples:

ACTIVATE(n__s(z0)) → c13(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__diff(z0, z1)) → c14(DIFF(activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))
ACTIVATE(n__p(z0)) → c15(P(activate(z0)), ACTIVATE(z0))

S tuples:

ACTIVATE(n__s(z0)) → c13(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__diff(z0, z1)) → c14(DIFF(activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))
ACTIVATE(n__p(z0)) → c15(P(activate(z0)), ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

p, leq, if, diff, 0, s, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c13, c14, c15

(5) 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.

ACTIVATE(n__p(z0)) → c15(P(activate(z0)), ACTIVATE(z0))

We considered the (Usable) Rules:

activate(n__0) → 0
activate(n__s(z0)) → s(activate(z0))
activate(n__diff(z0, z1)) → diff(activate(z0), activate(z1))
activate(n__p(z0)) → p(activate(z0))
activate(z0) → z0
p(z0) → n__p(z0)
diff(z0, z1) → if(leq(z0, z1), n__0, n__s(n__diff(n__p(z0), z1)))
diff(z0, z1) → n__diff(z0, z1)
s(z0) → n__s(z0)
0 → n__0

And the Tuples:

ACTIVATE(n__s(z0)) → c13(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__diff(z0, z1)) → c14(DIFF(activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))
ACTIVATE(n__p(z0)) → c15(P(activate(z0)), ACTIVATE(z0))

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

POL(0) = [1]   
POL(ACTIVATE(x₁)) = x₁   
POL(DIFF(x₁, x₂)) = 0   
POL(P(x₁)) = 0   
POL(S(x₁)) = 0   
POL(activate(x₁)) = [5]   
POL(c13(x₁, x₂)) = x₁ + x₂   
POL(c14(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c15(x₁, x₂)) = x₁ + x₂   
POL(diff(x₁, x₂)) = [3] + [3]x₁ + [2]x₂   
POL(if(x₁, x₂, x₃)) = [3] + [3]x₁ + [3]x₃   
POL(leq(x₁, x₂)) = [3] + [3]x₁ + [2]x₂   
POL(n__0) = [3]   
POL(n__diff(x₁, x₂)) = x₁ + x₂   
POL(n__p(x₁)) = [2] + x₁   
POL(n__s(x₁)) = x₁   
POL(p(x₁)) = [3] + [3]x₁   
POL(s(x₁)) = [2] + [3]x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(0) → 0
p(s(z0)) → z0
p(z0) → n__p(z0)
leq(0, z0) → true
leq(s(z0), 0) → false
leq(s(z0), s(z1)) → leq(z0, z1)
if(true, z0, z1) → activate(z0)
if(false, z0, z1) → activate(z1)
diff(z0, z1) → if(leq(z0, z1), n__0, n__s(n__diff(n__p(z0), z1)))
diff(z0, z1) → n__diff(z0, z1)
0 → n__0
s(z0) → n__s(z0)
activate(n__0) → 0
activate(n__s(z0)) → s(activate(z0))
activate(n__diff(z0, z1)) → diff(activate(z0), activate(z1))
activate(n__p(z0)) → p(activate(z0))
activate(z0) → z0

Tuples:

ACTIVATE(n__s(z0)) → c13(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__diff(z0, z1)) → c14(DIFF(activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))
ACTIVATE(n__p(z0)) → c15(P(activate(z0)), ACTIVATE(z0))

S tuples:

ACTIVATE(n__s(z0)) → c13(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__diff(z0, z1)) → c14(DIFF(activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))

K tuples:

ACTIVATE(n__p(z0)) → c15(P(activate(z0)), ACTIVATE(z0))

Defined Rule Symbols:

p, leq, if, diff, 0, s, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c13, c14, c15

(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.

ACTIVATE(n__s(z0)) → c13(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__diff(z0, z1)) → c14(DIFF(activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))

We considered the (Usable) Rules:

activate(n__0) → 0
activate(n__s(z0)) → s(activate(z0))
activate(n__diff(z0, z1)) → diff(activate(z0), activate(z1))
activate(n__p(z0)) → p(activate(z0))
activate(z0) → z0
p(z0) → n__p(z0)
diff(z0, z1) → if(leq(z0, z1), n__0, n__s(n__diff(n__p(z0), z1)))
diff(z0, z1) → n__diff(z0, z1)
s(z0) → n__s(z0)
0 → n__0

And the Tuples:

ACTIVATE(n__s(z0)) → c13(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__diff(z0, z1)) → c14(DIFF(activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))
ACTIVATE(n__p(z0)) → c15(P(activate(z0)), ACTIVATE(z0))

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

POL(0) = [3]   
POL(ACTIVATE(x₁)) = x₁   
POL(DIFF(x₁, x₂)) = 0   
POL(P(x₁)) = [1]   
POL(S(x₁)) = 0   
POL(activate(x₁)) = [3]   
POL(c13(x₁, x₂)) = x₁ + x₂   
POL(c14(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c15(x₁, x₂)) = x₁ + x₂   
POL(diff(x₁, x₂)) = [5] + [3]x₁ + [2]x₂   
POL(if(x₁, x₂, x₃)) = [3] + [2]x₁ + [3]x₂ + [3]x₃   
POL(leq(x₁, x₂)) = [2] + [3]x₁ + [4]x₂   
POL(n__0) = [2]   
POL(n__diff(x₁, x₂)) = [3] + x₁ + x₂   
POL(n__p(x₁)) = [4] + x₁   
POL(n__s(x₁)) = [2] + x₁   
POL(p(x₁)) = [3] + [4]x₁   
POL(s(x₁)) = [2] + [3]x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(0) → 0
p(s(z0)) → z0
p(z0) → n__p(z0)
leq(0, z0) → true
leq(s(z0), 0) → false
leq(s(z0), s(z1)) → leq(z0, z1)
if(true, z0, z1) → activate(z0)
if(false, z0, z1) → activate(z1)
diff(z0, z1) → if(leq(z0, z1), n__0, n__s(n__diff(n__p(z0), z1)))
diff(z0, z1) → n__diff(z0, z1)
0 → n__0
s(z0) → n__s(z0)
activate(n__0) → 0
activate(n__s(z0)) → s(activate(z0))
activate(n__diff(z0, z1)) → diff(activate(z0), activate(z1))
activate(n__p(z0)) → p(activate(z0))
activate(z0) → z0

Tuples:

ACTIVATE(n__s(z0)) → c13(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__diff(z0, z1)) → c14(DIFF(activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))
ACTIVATE(n__p(z0)) → c15(P(activate(z0)), ACTIVATE(z0))

S tuples:none
K tuples:

ACTIVATE(n__p(z0)) → c15(P(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c13(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__diff(z0, z1)) → c14(DIFF(activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))

Defined Rule Symbols:

p, leq, if, diff, 0, s, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c13, c14, c15

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(10) BOUNDS(O(1), O(1))