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:

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(x, z))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

implies(not(z0), z1) → or(z0, z1)
implies(not(z0), or(z1, z2)) → implies(z1, or(z0, z2))
implies(z0, or(z1, z2)) → or(z1, implies(z0, z2))

Tuples:

IMPLIES(not(z0), z1) → c
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))

S tuples:

IMPLIES(not(z0), z1) → c
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))

K tuples:none
Defined Rule Symbols:

implies

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c, c1, c2

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

Removed 1 trailing nodes:

IMPLIES(not(z0), z1) → c

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

implies(not(z0), z1) → or(z0, z1)
implies(not(z0), or(z1, z2)) → implies(z1, or(z0, z2))
implies(z0, or(z1, z2)) → or(z1, implies(z0, z2))

Tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))

S tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))

K tuples:none
Defined Rule Symbols:

implies

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c1, c2

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

implies(not(z0), z1) → or(z0, z1)
implies(not(z0), or(z1, z2)) → implies(z1, or(z0, z2))
implies(z0, or(z1, z2)) → or(z1, implies(z0, z2))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))

S tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c1, c2

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

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

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))

We considered the (Usable) Rules:none
And the Tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))

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

POL(IMPLIES(x₁, x₂)) = [4]x₁ + [4]x₂   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(not(x₁)) = [4] + x₁   
POL(or(x₁, x₂)) = [4] + x₁ + x₂   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))

S tuples:none
K tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))

Defined Rule Symbols:none

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c1, c2

(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

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