(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(x))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
w(r(z0)) → r(w(z0))
b(r(z0)) → r(b(z0))
b(w(z0)) → w(b(z0))
Tuples:
W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))
S tuples:
W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))
K tuples:none
Defined Rule Symbols:
w, b
Defined Pair Symbols:
W, B
Compound Symbols:
c, c1, c2
(3) 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.
B(r(z0)) → c1(B(z0))
We considered the (Usable) Rules:
b(w(z0)) → w(b(z0))
w(r(z0)) → r(w(z0))
And the Tuples:
W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(B(x1)) = [2]x1
POL(W(x1)) = [3]
POL(b(x1)) = [1] + [5]x1
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(r(x1)) = [3] + x1
POL(w(x1)) = [4] + [4]x1
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
w(r(z0)) → r(w(z0))
b(r(z0)) → r(b(z0))
b(w(z0)) → w(b(z0))
Tuples:
W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))
S tuples:
W(r(z0)) → c(W(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))
K tuples:
B(r(z0)) → c1(B(z0))
Defined Rule Symbols:
w, b
Defined Pair Symbols:
W, B
Compound Symbols:
c, c1, c2
(5) 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.
B(w(z0)) → c2(W(b(z0)), B(z0))
We considered the (Usable) Rules:
b(w(z0)) → w(b(z0))
w(r(z0)) → r(w(z0))
And the Tuples:
W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(B(x1)) = x12
POL(W(x1)) = 0
POL(b(x1)) = 0
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(r(x1)) = x1
POL(w(x1)) = [2] + x1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
w(r(z0)) → r(w(z0))
b(r(z0)) → r(b(z0))
b(w(z0)) → w(b(z0))
Tuples:
W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))
S tuples:
W(r(z0)) → c(W(z0))
K tuples:
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))
Defined Rule Symbols:
w, b
Defined Pair Symbols:
W, B
Compound Symbols:
c, c1, c2
(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.
W(r(z0)) → c(W(z0))
We considered the (Usable) Rules:
b(w(z0)) → w(b(z0))
w(r(z0)) → r(w(z0))
And the Tuples:
W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(B(x1)) = [4]x1
POL(W(x1)) = [5] + [2]x1
POL(b(x1)) = [1] + [4]x1
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(r(x1)) = [1] + x1
POL(w(x1)) = [4] + [4]x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
w(r(z0)) → r(w(z0))
b(r(z0)) → r(b(z0))
b(w(z0)) → w(b(z0))
Tuples:
W(r(z0)) → c(W(z0))
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))
S tuples:none
K tuples:
B(r(z0)) → c1(B(z0))
B(w(z0)) → c2(W(b(z0)), B(z0))
W(r(z0)) → c(W(z0))
Defined Rule Symbols:
w, b
Defined Pair Symbols:
W, B
Compound Symbols:
c, c1, c2
(9) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(10) BOUNDS(O(1), O(1))