(0) Obligation:

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

and(x, false) → false
and(x, not(false)) → x
not(not(x)) → x
implies(false, y) → not(false)
implies(x, false) → not(x)
implies(not(x), not(y)) → implies(y, and(x, y))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(z0, false) → false
and(z0, not(false)) → z0
not(not(z0)) → z0
implies(false, z0) → not(false)
implies(z0, false) → not(z0)
implies(not(z0), not(z1)) → implies(z1, and(z0, z1))
Tuples:

IMPLIES(false, z0) → c3(NOT(false))
IMPLIES(z0, false) → c4(NOT(z0))
IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))
S tuples:

IMPLIES(false, z0) → c3(NOT(false))
IMPLIES(z0, false) → c4(NOT(z0))
IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))
K tuples:none
Defined Rule Symbols:

and, not, implies

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c3, c4, c5

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

Removed 2 trailing nodes:

IMPLIES(false, z0) → c3(NOT(false))
IMPLIES(z0, false) → c4(NOT(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(z0, false) → false
and(z0, not(false)) → z0
not(not(z0)) → z0
implies(false, z0) → not(false)
implies(z0, false) → not(z0)
implies(not(z0), not(z1)) → implies(z1, and(z0, z1))
Tuples:

IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))
S tuples:

IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))
K tuples:none
Defined Rule Symbols:

and, not, implies

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c5

(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.

IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))
We considered the (Usable) Rules:

and(z0, false) → false
And the Tuples:

IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(AND(x1, x2)) = [5] + x2   
POL(IMPLIES(x1, x2)) = [5]x1 + [2]x2   
POL(and(x1, x2)) = [5] + [4]x1 + x2   
POL(c5(x1, x2)) = x1 + x2   
POL(false) = 0   
POL(not(x1)) = [5] + [4]x1   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(z0, false) → false
and(z0, not(false)) → z0
not(not(z0)) → z0
implies(false, z0) → not(false)
implies(z0, false) → not(z0)
implies(not(z0), not(z1)) → implies(z1, and(z0, z1))
Tuples:

IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))
S tuples:none
K tuples:

IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))
Defined Rule Symbols:

and, not, implies

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c5

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))