(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → z
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(z0, z1, a) → f(b(b(f(z0), z0), z1))
b(z0, b(z1, a)) → f(b(c(f(a), z0, z1), z1))
f(c(c(z0, a, a), z1, a)) → z0
Tuples:
C(z0, z1, a) → c1(F(b(b(f(z0), z0), z1)), B(b(f(z0), z0), z1), B(f(z0), z0), F(z0))
B(z0, b(z1, a)) → c2(F(b(c(f(a), z0, z1), z1)), B(c(f(a), z0, z1), z1), C(f(a), z0, z1), F(a))
S tuples:
C(z0, z1, a) → c1(F(b(b(f(z0), z0), z1)), B(b(f(z0), z0), z1), B(f(z0), z0), F(z0))
B(z0, b(z1, a)) → c2(F(b(c(f(a), z0, z1), z1)), B(c(f(a), z0, z1), z1), C(f(a), z0, z1), F(a))
K tuples:none
Defined Rule Symbols:
c, b, f
Defined Pair Symbols:
C, B
Compound Symbols:
c1, c2
(3) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 4 trailing tuple parts
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(z0, z1, a) → f(b(b(f(z0), z0), z1))
b(z0, b(z1, a)) → f(b(c(f(a), z0, z1), z1))
f(c(c(z0, a, a), z1, a)) → z0
Tuples:
C(z0, z1, a) → c1(B(b(f(z0), z0), z1), B(f(z0), z0))
B(z0, b(z1, a)) → c2(B(c(f(a), z0, z1), z1), C(f(a), z0, z1))
S tuples:
C(z0, z1, a) → c1(B(b(f(z0), z0), z1), B(f(z0), z0))
B(z0, b(z1, a)) → c2(B(c(f(a), z0, z1), z1), C(f(a), z0, z1))
K tuples:none
Defined Rule Symbols:
c, b, f
Defined Pair Symbols:
C, B
Compound Symbols:
c1, c2
(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.
C(z0, z1, a) → c1(B(b(f(z0), z0), z1), B(f(z0), z0))
We considered the (Usable) Rules:
c(z0, z1, a) → f(b(b(f(z0), z0), z1))
b(z0, b(z1, a)) → f(b(c(f(a), z0, z1), z1))
And the Tuples:
C(z0, z1, a) → c1(B(b(f(z0), z0), z1), B(f(z0), z0))
B(z0, b(z1, a)) → c2(B(c(f(a), z0, z1), z1), C(f(a), z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(B(x1, x2)) = [4]x1 + x2
POL(C(x1, x2, x3)) = [2] + [5]x1 + [4]x2
POL(a) = [2]
POL(b(x1, x2)) = [4]x1 + x2
POL(c(x1, x2, x3)) = 0
POL(c1(x1, x2)) = x1 + x2
POL(c2(x1, x2)) = x1 + x2
POL(f(x1)) = 0
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(z0, z1, a) → f(b(b(f(z0), z0), z1))
b(z0, b(z1, a)) → f(b(c(f(a), z0, z1), z1))
f(c(c(z0, a, a), z1, a)) → z0
Tuples:
C(z0, z1, a) → c1(B(b(f(z0), z0), z1), B(f(z0), z0))
B(z0, b(z1, a)) → c2(B(c(f(a), z0, z1), z1), C(f(a), z0, z1))
S tuples:
B(z0, b(z1, a)) → c2(B(c(f(a), z0, z1), z1), C(f(a), z0, z1))
K tuples:
C(z0, z1, a) → c1(B(b(f(z0), z0), z1), B(f(z0), z0))
Defined Rule Symbols:
c, b, f
Defined Pair Symbols:
C, B
Compound Symbols:
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.
B(z0, b(z1, a)) → c2(B(c(f(a), z0, z1), z1), C(f(a), z0, z1))
We considered the (Usable) Rules:
c(z0, z1, a) → f(b(b(f(z0), z0), z1))
b(z0, b(z1, a)) → f(b(c(f(a), z0, z1), z1))
And the Tuples:
C(z0, z1, a) → c1(B(b(f(z0), z0), z1), B(f(z0), z0))
B(z0, b(z1, a)) → c2(B(c(f(a), z0, z1), z1), C(f(a), z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(B(x1, x2)) = [1] + [2]x1 + [2]x2
POL(C(x1, x2, x3)) = [2] + [5]x1 + [2]x2 + [2]x3
POL(a) = [2]
POL(b(x1, x2)) = [2] + [4]x1
POL(c(x1, x2, x3)) = 0
POL(c1(x1, x2)) = x1 + x2
POL(c2(x1, x2)) = x1 + x2
POL(f(x1)) = 0
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(z0, z1, a) → f(b(b(f(z0), z0), z1))
b(z0, b(z1, a)) → f(b(c(f(a), z0, z1), z1))
f(c(c(z0, a, a), z1, a)) → z0
Tuples:
C(z0, z1, a) → c1(B(b(f(z0), z0), z1), B(f(z0), z0))
B(z0, b(z1, a)) → c2(B(c(f(a), z0, z1), z1), C(f(a), z0, z1))
S tuples:none
K tuples:
C(z0, z1, a) → c1(B(b(f(z0), z0), z1), B(f(z0), z0))
B(z0, b(z1, a)) → c2(B(c(f(a), z0, z1), z1), C(f(a), z0, z1))
Defined Rule Symbols:
c, b, f
Defined Pair Symbols:
C, B
Compound Symbols:
c1, c2
(9) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(10) BOUNDS(O(1), O(1))