(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:
+(+(x, y), z) → +(x, +(y, z))
+(f(x), f(y)) → f(+(x, y))
+(f(x), +(f(y), z)) → +(f(+(x, y)), 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:
+(+(x, y), z) → +(x, +(y, z))
+(f(x), +(f(y), z)) → +(f(+(x, y)), z)
(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(x), f(y)) → f(+(x, y))
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]
transitions:
f0(0) → 0
+0(0, 0) → 1
+1(0, 0) → 2
f1(2) → 1
f1(2) → 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:
+(f(z0), f(z1)) → f(+(z0, z1))
Tuples:
+'(f(z0), f(z1)) → c(+'(z0, z1))
S tuples:
+'(f(z0), f(z1)) → c(+'(z0, z1))
K tuples:none
Defined Rule Symbols:
+
Defined Pair Symbols:
+'
Compound Symbols:
c
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
+(f(z0), f(z1)) → f(+(z0, z1))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
+'(f(z0), f(z1)) → c(+'(z0, z1))
S tuples:
+'(f(z0), f(z1)) → c(+'(z0, z1))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
+'
Compound Symbols:
c
(9) 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), f(z1)) → c(+'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
+'(f(z0), f(z1)) → c(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(+'(x1, x2)) = x2
POL(c(x1)) = x1
POL(f(x1)) = [1] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
+'(f(z0), f(z1)) → c(+'(z0, z1))
S tuples:none
K tuples:
+'(f(z0), f(z1)) → c(+'(z0, z1))
Defined Rule Symbols:none
Defined Pair Symbols:
+'
Compound Symbols:
c
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)