(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
U11(tt, M, N) → U12(tt, M, N)
U12(tt, M, N) → s(plus(N, M))
U21(tt, M, N) → U22(tt, M, N)
U22(tt, M, N) → plus(x(N, M), N)
plus(N, 0) → N
plus(N, s(M)) → U11(tt, M, N)
x(N, 0) → 0
x(N, s(M)) → U21(tt, M, N)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
U11(tt, z0, z1) → U12(tt, z0, z1)
U12(tt, z0, z1) → s(plus(z1, z0))
U21(tt, z0, z1) → U22(tt, z0, z1)
U22(tt, z0, z1) → plus(x(z1, z0), z1)
plus(z0, 0) → z0
plus(z0, s(z1)) → U11(tt, z1, z0)
x(z0, 0) → 0
x(z0, s(z1)) → U21(tt, z1, z0)
Tuples:
U11'(tt, z0, z1) → c(U12'(tt, z0, z1))
U12'(tt, z0, z1) → c1(PLUS(z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, z0, z1))
U22'(tt, z0, z1) → c3(PLUS(x(z1, z0), z1), X(z1, z0))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
S tuples:
U11'(tt, z0, z1) → c(U12'(tt, z0, z1))
U12'(tt, z0, z1) → c1(PLUS(z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, z0, z1))
U22'(tt, z0, z1) → c3(PLUS(x(z1, z0), z1), X(z1, z0))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
K tuples:none
Defined Rule Symbols:
U11, U12, U21, U22, plus, x
Defined Pair Symbols:
U11', U12', U21', U22', PLUS, X
Compound Symbols:
c, c1, c2, c3, c5, c7
(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.
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
We considered the (Usable) Rules:
x(z0, 0) → 0
x(z0, s(z1)) → U21(tt, z1, z0)
U21(tt, z0, z1) → U22(tt, z0, z1)
U22(tt, z0, z1) → plus(x(z1, z0), z1)
plus(z0, 0) → z0
plus(z0, s(z1)) → U11(tt, z1, z0)
U11(tt, z0, z1) → U12(tt, z0, z1)
U12(tt, z0, z1) → s(plus(z1, z0))
And the Tuples:
U11'(tt, z0, z1) → c(U12'(tt, z0, z1))
U12'(tt, z0, z1) → c1(PLUS(z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, z0, z1))
U22'(tt, z0, z1) → c3(PLUS(x(z1, z0), z1), X(z1, z0))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [1]
POL(PLUS(x1, x2)) = 0
POL(U11(x1, x2, x3)) = [4] + [3]x1 + x2 + [3]x3
POL(U11'(x1, x2, x3)) = [4]x1
POL(U12(x1, x2, x3)) = [3] + [4]x1 + [3]x2 + [5]x3
POL(U12'(x1, x2, x3)) = 0
POL(U21(x1, x2, x3)) = [5]x1 + [3]x2 + [3]x3
POL(U21'(x1, x2, x3)) = [3] + x1 + [2]x2
POL(U22(x1, x2, x3)) = [2] + [2]x1 + [2]x2 + [3]x3
POL(U22'(x1, x2, x3)) = [3] + [2]x1 + [2]x2
POL(X(x1, x2)) = [3] + [2]x2
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c7(x1)) = x1
POL(plus(x1, x2)) = [2] + [3]x1 + [3]x2
POL(s(x1)) = [4] + x1
POL(tt) = 0
POL(x(x1, x2)) = [3] + [3]x1 + [2]x2
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
U11(tt, z0, z1) → U12(tt, z0, z1)
U12(tt, z0, z1) → s(plus(z1, z0))
U21(tt, z0, z1) → U22(tt, z0, z1)
U22(tt, z0, z1) → plus(x(z1, z0), z1)
plus(z0, 0) → z0
plus(z0, s(z1)) → U11(tt, z1, z0)
x(z0, 0) → 0
x(z0, s(z1)) → U21(tt, z1, z0)
Tuples:
U11'(tt, z0, z1) → c(U12'(tt, z0, z1))
U12'(tt, z0, z1) → c1(PLUS(z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, z0, z1))
U22'(tt, z0, z1) → c3(PLUS(x(z1, z0), z1), X(z1, z0))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
S tuples:
U11'(tt, z0, z1) → c(U12'(tt, z0, z1))
U12'(tt, z0, z1) → c1(PLUS(z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, z0, z1))
U22'(tt, z0, z1) → c3(PLUS(x(z1, z0), z1), X(z1, z0))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
K tuples:
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
Defined Rule Symbols:
U11, U12, U21, U22, plus, x
Defined Pair Symbols:
U11', U12', U21', U22', PLUS, X
Compound Symbols:
c, c1, c2, c3, c5, c7
(5) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
U21'(tt, z0, z1) → c2(U22'(tt, z0, z1))
U22'(tt, z0, z1) → c3(PLUS(x(z1, z0), z1), X(z1, z0))
U22'(tt, z0, z1) → c3(PLUS(x(z1, z0), z1), X(z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
U11(tt, z0, z1) → U12(tt, z0, z1)
U12(tt, z0, z1) → s(plus(z1, z0))
U21(tt, z0, z1) → U22(tt, z0, z1)
U22(tt, z0, z1) → plus(x(z1, z0), z1)
plus(z0, 0) → z0
plus(z0, s(z1)) → U11(tt, z1, z0)
x(z0, 0) → 0
x(z0, s(z1)) → U21(tt, z1, z0)
Tuples:
U11'(tt, z0, z1) → c(U12'(tt, z0, z1))
U12'(tt, z0, z1) → c1(PLUS(z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, z0, z1))
U22'(tt, z0, z1) → c3(PLUS(x(z1, z0), z1), X(z1, z0))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
S tuples:
U11'(tt, z0, z1) → c(U12'(tt, z0, z1))
U12'(tt, z0, z1) → c1(PLUS(z1, z0))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
K tuples:
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, z0, z1))
U22'(tt, z0, z1) → c3(PLUS(x(z1, z0), z1), X(z1, z0))
Defined Rule Symbols:
U11, U12, U21, U22, plus, x
Defined Pair Symbols:
U11', U12', U21', U22', PLUS, X
Compound Symbols:
c, c1, c2, c3, c5, c7
(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.
U12'(tt, z0, z1) → c1(PLUS(z1, z0))
We considered the (Usable) Rules:
x(z0, 0) → 0
x(z0, s(z1)) → U21(tt, z1, z0)
U21(tt, z0, z1) → U22(tt, z0, z1)
U22(tt, z0, z1) → plus(x(z1, z0), z1)
plus(z0, 0) → z0
plus(z0, s(z1)) → U11(tt, z1, z0)
U11(tt, z0, z1) → U12(tt, z0, z1)
U12(tt, z0, z1) → s(plus(z1, z0))
And the Tuples:
U11'(tt, z0, z1) → c(U12'(tt, z0, z1))
U12'(tt, z0, z1) → c1(PLUS(z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, z0, z1))
U22'(tt, z0, z1) → c3(PLUS(x(z1, z0), z1), X(z1, z0))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(PLUS(x1, x2)) = x2
POL(U11(x1, x2, x3)) = 0
POL(U11'(x1, x2, x3)) = [1] + x2
POL(U12(x1, x2, x3)) = 0
POL(U12'(x1, x2, x3)) = [1] + x2
POL(U21(x1, x2, x3)) = 0
POL(U21'(x1, x2, x3)) = x3 + x2·x3
POL(U22(x1, x2, x3)) = 0
POL(U22'(x1, x2, x3)) = x3 + x2·x3
POL(X(x1, x2)) = x1·x2
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c7(x1)) = x1
POL(plus(x1, x2)) = 0
POL(s(x1)) = [1] + x1
POL(tt) = 0
POL(x(x1, x2)) = 0
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
U11(tt, z0, z1) → U12(tt, z0, z1)
U12(tt, z0, z1) → s(plus(z1, z0))
U21(tt, z0, z1) → U22(tt, z0, z1)
U22(tt, z0, z1) → plus(x(z1, z0), z1)
plus(z0, 0) → z0
plus(z0, s(z1)) → U11(tt, z1, z0)
x(z0, 0) → 0
x(z0, s(z1)) → U21(tt, z1, z0)
Tuples:
U11'(tt, z0, z1) → c(U12'(tt, z0, z1))
U12'(tt, z0, z1) → c1(PLUS(z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, z0, z1))
U22'(tt, z0, z1) → c3(PLUS(x(z1, z0), z1), X(z1, z0))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
S tuples:
U11'(tt, z0, z1) → c(U12'(tt, z0, z1))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
K tuples:
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, z0, z1))
U22'(tt, z0, z1) → c3(PLUS(x(z1, z0), z1), X(z1, z0))
U12'(tt, z0, z1) → c1(PLUS(z1, z0))
Defined Rule Symbols:
U11, U12, U21, U22, plus, x
Defined Pair Symbols:
U11', U12', U21', U22', PLUS, X
Compound Symbols:
c, c1, c2, c3, c5, c7
(9) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
U11'(tt, z0, z1) → c(U12'(tt, z0, z1))
U12'(tt, z0, z1) → c1(PLUS(z1, z0))
Now S is empty
(10) BOUNDS(O(1), O(1))