(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, n^1).
The TRS R consists of the following rules:
f(0, y) → 0
f(s(x), y) → f(f(x, y), y)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0, z0) → 0
f(s(z0), z1) → f(f(z0, z1), z1)
Tuples:
F(0, z0) → c
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:
F(0, z0) → c
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c, c1
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
F(0, z0) → c
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0, z0) → 0
f(s(z0), z1) → f(f(z0, z1), z1)
Tuples:
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1
(5) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
We considered the (Usable) Rules:
f(0, z0) → 0
f(s(z0), z1) → f(f(z0, z1), z1)
And the Tuples:
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(F(x1, x2)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(f(x1, x2)) = 0
POL(s(x1)) = [1] + x1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0, z0) → 0
f(s(z0), z1) → f(f(z0, z1), z1)
Tuples:
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:none
K tuples:
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1
(7) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(8) BOUNDS(1, 1)