(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
:(:(z0, z1), z2) → :(z0, :(z1, z2))
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2))
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:
:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
:'(z0, +(z1, f(z2))) → c2(:'(g(z0, z2), +(z1, a)))
S tuples:
:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
:'(z0, +(z1, f(z2))) → c2(:'(g(z0, z2), +(z1, a)))
K tuples:none
Defined Rule Symbols:
:
Defined Pair Symbols:
:'
Compound Symbols:
c, c1, c2
(3) CdtGraphRemoveDanglingProof (ComplexityIfPolyImplication transformation)
Removed 1 of 3 dangling nodes:
:'(z0, +(z1, f(z2))) → c2(:'(g(z0, z2), +(z1, a)))
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
:(:(z0, z1), z2) → :(z0, :(z1, z2))
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2))
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:
:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
S tuples:
:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
K tuples:none
Defined Rule Symbols:
:
Defined Pair Symbols:
:'
Compound Symbols:
c, c1
(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.
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
We considered the (Usable) Rules:
:(:(z0, z1), z2) → :(z0, :(z1, z2))
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2))
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
And the Tuples:
:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(+(x1, x2)) = [4] + x1 + x2
POL(:(x1, x2)) = [4] + x1 + [2]x2
POL(:'(x1, x2)) = [2]x1
POL(a) = [3]
POL(c(x1, x2)) = x1 + x2
POL(c1(x1, x2)) = x1 + x2
POL(f(x1)) = [4]
POL(g(x1, x2)) = [5]
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
:(:(z0, z1), z2) → :(z0, :(z1, z2))
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2))
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:
:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
S tuples:
:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
K tuples:
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
Defined Rule Symbols:
:
Defined Pair Symbols:
:'
Compound Symbols:
c, c1
(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.
:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
We considered the (Usable) Rules:
:(:(z0, z1), z2) → :(z0, :(z1, z2))
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2))
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
And the Tuples:
:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(+(x1, x2)) = [4] + x1 + x2
POL(:(x1, x2)) = [4] + [4]x1 + x2
POL(:'(x1, x2)) = [3] + [2]x1
POL(a) = [3]
POL(c(x1, x2)) = x1 + x2
POL(c1(x1, x2)) = x1 + x2
POL(f(x1)) = [3]
POL(g(x1, x2)) = [2]
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
:(:(z0, z1), z2) → :(z0, :(z1, z2))
:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2))
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:
:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
S tuples:none
K tuples:
:'(+(z0, z1), z2) → c1(:'(z0, z2), :'(z1, z2))
:'(:(z0, z1), z2) → c(:'(z0, :(z1, z2)), :'(z1, z2))
Defined Rule Symbols:
:
Defined Pair Symbols:
:'
Compound Symbols:
c, c1
(9) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(10) BOUNDS(O(1), O(1))