(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, 0) → s(0)
f(s(x), s(y)) → s(f(x, y))
g(0, x) → g(f(x, x), x)
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, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
g(0, z0) → g(f(z0, z0), z0)
Tuples:
F(z0, 0) → c
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
F(z0, 0) → c
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c, c1, c2
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
F(z0, 0) → c
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
g(0, z0) → g(f(z0, z0), z0)
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
g(0, z0) → g(f(z0, z0), z0)
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2
(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.
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
We considered the (Usable) Rules:
f(s(z0), s(z1)) → s(f(z0, z1))
f(z0, 0) → s(0)
And the Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [3]
POL(F(x1, x2)) = [2]
POL(G(x1, x2)) = [3]x1
POL(c1(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(f(x1, x2)) = [1]
POL(s(x1)) = 0
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
K tuples:
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
Defined Rule Symbols:
f
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2
(9) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
G(
0,
z0) →
c2(
G(
f(
z0,
z0),
z0),
F(
z0,
z0)) by
G(0, 0) → c2(G(s(0), 0), F(0, 0))
G(0, s(z0)) → c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, 0) → c2(G(s(0), 0), F(0, 0))
G(0, s(z0)) → c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
K tuples:
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
Defined Rule Symbols:
f
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2
(11) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
G(0, 0) → c2(G(s(0), 0), F(0, 0))
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, s(z0)) → c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2
(13) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, s(z0)) → c2(F(s(z0), s(z0)))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2
(15) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
G(0, s(z0)) → c2(F(s(z0), s(z0)))
(16) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1
(17) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
(18) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c1
(19) 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), s(z1)) → c1(F(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1, x2)) = x2
POL(c1(x1)) = x1
POL(s(x1)) = [1] + x1
(20) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
S tuples:none
K tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c1
(21) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(22) BOUNDS(1, 1)