(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(x, g(x)) → x
f(x, h(y)) → f(h(x), 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(z0, g(z0)) → z0
f(z0, h(z1)) → f(h(z0), z1)
Tuples:
F(z0, g(z0)) → c
F(z0, h(z1)) → c1(F(h(z0), z1))
S tuples:
F(z0, g(z0)) → c
F(z0, h(z1)) → c1(F(h(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(z0, g(z0)) → c
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, g(z0)) → z0
f(z0, h(z1)) → f(h(z0), z1)
Tuples:
F(z0, h(z1)) → c1(F(h(z0), z1))
S tuples:
F(z0, h(z1)) → c1(F(h(z0), z1))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
f(z0, g(z0)) → z0
f(z0, h(z1)) → f(h(z0), z1)
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(z0, h(z1)) → c1(F(h(z0), z1))
S tuples:
F(z0, h(z1)) → c1(F(h(z0), z1))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c1
(7) 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(z0, h(z1)) → c1(F(h(z0), z1))
We considered the (Usable) Rules:none
And the Tuples:
F(z0, h(z1)) → c1(F(h(z0), z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1, x2)) = x2
POL(c1(x1)) = x1
POL(h(x1)) = [1] + x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(z0, h(z1)) → c1(F(h(z0), z1))
S tuples:none
K tuples:
F(z0, h(z1)) → c1(F(h(z0), z1))
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c1
(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(10) BOUNDS(1, 1)