(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(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(e, x)
Rewrite Strategy: INNERMOST
(1) NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID) transformation)
The following defined symbols can occur below the 0th argument of g: f, g
The following defined symbols can occur below the 1th argument of g: f, g
Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
f(g(x, y)) → g(f(x), f(y))
(2) 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(a) → b
f(c) → d
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(e, x)
Rewrite Strategy: INNERMOST
(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(a) → b
f(c) → d
f(h(z0, z1)) → g(h(z1, f(z0)), h(z0, f(z1)))
g(z0, z0) → h(e, z0)
Tuples:
F(a) → c1
F(c) → c2
F(h(z0, z1)) → c3(G(h(z1, f(z0)), h(z0, f(z1))), F(z0), F(z1))
G(z0, z0) → c4
S tuples:
F(a) → c1
F(c) → c2
F(h(z0, z1)) → c3(G(h(z1, f(z0)), h(z0, f(z1))), F(z0), F(z1))
G(z0, z0) → c4
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2, c3, c4
(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 3 trailing nodes:
G(z0, z0) → c4
F(a) → c1
F(c) → c2
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(a) → b
f(c) → d
f(h(z0, z1)) → g(h(z1, f(z0)), h(z0, f(z1)))
g(z0, z0) → h(e, z0)
Tuples:
F(h(z0, z1)) → c3(G(h(z1, f(z0)), h(z0, f(z1))), F(z0), F(z1))
S tuples:
F(h(z0, z1)) → c3(G(h(z1, f(z0)), h(z0, f(z1))), F(z0), F(z1))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F
Compound Symbols:
c3
(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(a) → b
f(c) → d
f(h(z0, z1)) → g(h(z1, f(z0)), h(z0, f(z1)))
g(z0, z0) → h(e, z0)
Tuples:
F(h(z0, z1)) → c3(F(z0), F(z1))
S tuples:
F(h(z0, z1)) → c3(F(z0), F(z1))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F
Compound Symbols:
c3
(9) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
f(a) → b
f(c) → d
f(h(z0, z1)) → g(h(z1, f(z0)), h(z0, f(z1)))
g(z0, z0) → h(e, z0)
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(h(z0, z1)) → c3(F(z0), F(z1))
S tuples:
F(h(z0, z1)) → c3(F(z0), F(z1))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c3
(11) 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(h(z0, z1)) → c3(F(z0), F(z1))
We considered the (Usable) Rules:none
And the Tuples:
F(h(z0, z1)) → c3(F(z0), F(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1)) = [2] + [2]x1
POL(c3(x1, x2)) = x1 + x2
POL(h(x1, x2)) = [2] + x1 + x2
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(h(z0, z1)) → c3(F(z0), F(z1))
S tuples:none
K tuples:
F(h(z0, z1)) → c3(F(z0), F(z1))
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c3
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(14) BOUNDS(1, 1)