(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
and(tt, X) → activate(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
activate(X) → X
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(tt, z0) → activate(z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
x(z0, 0) → 0
x(z0, s(z1)) → plus(x(z0, z1), z0)
activate(z0) → z0
Tuples:
AND(tt, z0) → c(ACTIVATE(z0))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
S tuples:
AND(tt, z0) → c(ACTIVATE(z0))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
K tuples:none
Defined Rule Symbols:
and, plus, x, activate
Defined Pair Symbols:
AND, PLUS, X
Compound Symbols:
c, c2, c4
(3) CdtGraphRemoveDanglingProof (ComplexityIfPolyImplication transformation)
Removed 1 of 3 dangling nodes:
AND(tt, z0) → c(ACTIVATE(z0))
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(tt, z0) → activate(z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
x(z0, 0) → 0
x(z0, s(z1)) → plus(x(z0, z1), z0)
activate(z0) → z0
Tuples:
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
S tuples:
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
K tuples:none
Defined Rule Symbols:
and, plus, x, activate
Defined Pair Symbols:
PLUS, X
Compound Symbols:
c2, c4
(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.
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
We considered the (Usable) Rules:
x(z0, 0) → 0
x(z0, s(z1)) → plus(x(z0, z1), z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
And the Tuples:
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [3]
POL(PLUS(x1, x2)) = 0
POL(X(x1, x2)) = x2
POL(c2(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(plus(x1, x2)) = [2] + [2]x1
POL(s(x1)) = [2] + x1
POL(x(x1, x2)) = [4] + [2]x2
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(tt, z0) → activate(z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
x(z0, 0) → 0
x(z0, s(z1)) → plus(x(z0, z1), z0)
activate(z0) → z0
Tuples:
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
S tuples:
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
K tuples:
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
Defined Rule Symbols:
and, plus, x, activate
Defined Pair Symbols:
PLUS, X
Compound Symbols:
c2, c4
(7) 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.
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
We considered the (Usable) Rules:
x(z0, 0) → 0
x(z0, s(z1)) → plus(x(z0, z1), z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
And the Tuples:
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(PLUS(x1, x2)) = [1] + x2
POL(X(x1, x2)) = [2]x2 + [2]x1·x2
POL(c2(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(plus(x1, x2)) = 0
POL(s(x1)) = [1] + x1
POL(x(x1, x2)) = 0
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(tt, z0) → activate(z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
x(z0, 0) → 0
x(z0, s(z1)) → plus(x(z0, z1), z0)
activate(z0) → z0
Tuples:
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
S tuples:none
K tuples:
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
Defined Rule Symbols:
and, plus, x, activate
Defined Pair Symbols:
PLUS, X
Compound Symbols:
c2, c4
(9) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(10) BOUNDS(O(1), O(1))