(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
g(x, y) → x
g(x, y) → y
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
g(z0, z1) → z0
g(z0, z1) → z1
f(0, 1, z0) → f(s(z0), z0, z0)
f(z0, z1, s(z2)) → s(f(0, 1, z2))
Tuples:
G(z0, z1) → c
G(z0, z1) → c1
F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
S tuples:
G(z0, z1) → c
G(z0, z1) → c1
F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:
g, f
Defined Pair Symbols:
G, F
Compound Symbols:
c, c1, c2, c3
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
G(z0, z1) → c1
G(z0, z1) → c
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
g(z0, z1) → z0
g(z0, z1) → z1
f(0, 1, z0) → f(s(z0), z0, z0)
f(z0, z1, s(z2)) → s(f(0, 1, z2))
Tuples:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
S tuples:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:
g, f
Defined Pair Symbols:
F
Compound Symbols:
c2, c3
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
g(z0, z1) → z0
g(z0, z1) → z1
f(0, 1, z0) → f(s(z0), z0, z0)
f(z0, z1, s(z2)) → s(f(0, 1, z2))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
S tuples:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c2, c3
(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, z1, s(z2)) → c3(F(0, 1, z2))
We considered the (Usable) Rules:none
And the Tuples:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [4]
POL(1) = 0
POL(F(x1, x2, x3)) = [4]x3
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(s(x1)) = [4] + x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
S tuples:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
K tuples:
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c2, c3
(9) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
Now S is empty
(10) BOUNDS(1, 1)