(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, 1).
The TRS R consists of the following rules:
f(c(a, z, x)) → b(a, z)
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))
b(y, z) → z
Rewrite Strategy: INNERMOST
(1) NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID) transformation)
The TRS does not nest defined symbols.
Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))
(2) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, 1).
The TRS R consists of the following rules:
b(y, z) → z
f(c(a, z, x)) → b(a, z)
Rewrite Strategy: INNERMOST
(3) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)
A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 1.
The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2]
transitions:
c0(0, 0, 0) → 0
a0() → 0
b0(0, 0) → 1
f0(0) → 2
a1() → 3
b1(3, 0) → 2
0 → 1
0 → 2
(4) BOUNDS(1, n^1)
(5) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
b(z0, z1) → z1
f(c(a, z0, z1)) → b(a, z0)
Tuples:
B(z0, z1) → c1
F(c(a, z0, z1)) → c2(B(a, z0))
S tuples:
B(z0, z1) → c1
F(c(a, z0, z1)) → c2(B(a, z0))
K tuples:none
Defined Rule Symbols:
b, f
Defined Pair Symbols:
B, F
Compound Symbols:
c1, c2
(7) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
B(z0, z1) → c1
F(c(a, z0, z1)) → c2(B(a, z0))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
b(z0, z1) → z1
f(c(a, z0, z1)) → b(a, z0)
Tuples:none
S tuples:none
K tuples:none
Defined Rule Symbols:
b, f
Defined Pair Symbols:none
Compound Symbols:none
(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(10) BOUNDS(1, 1)