(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:
append(Cons(x, xs), ys) → Cons(x, append(xs, ys))
append(Nil, ys) → ys
goal(x, y) → append(x, y)
Rewrite Strategy: INNERMOST
(1) 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:
Cons0(0, 0) → 0
Nil0() → 0
append0(0, 0) → 1
goal0(0, 0) → 2
append1(0, 0) → 3
Cons1(0, 3) → 1
append1(0, 0) → 2
Cons1(0, 3) → 2
Cons1(0, 3) → 3
0 → 1
0 → 2
0 → 3
(2) BOUNDS(1, n^1)
(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
append(Cons(z0, z1), z2) → Cons(z0, append(z1, z2))
append(Nil, z0) → z0
goal(z0, z1) → append(z0, z1)
Tuples:
APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))
APPEND(Nil, z0) → c1
GOAL(z0, z1) → c2(APPEND(z0, z1))
S tuples:
APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))
APPEND(Nil, z0) → c1
GOAL(z0, z1) → c2(APPEND(z0, z1))
K tuples:none
Defined Rule Symbols:
append, goal
Defined Pair Symbols:
APPEND, GOAL
Compound Symbols:
c, c1, c2
(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
GOAL(z0, z1) → c2(APPEND(z0, z1))
Removed 1 trailing nodes:
APPEND(Nil, z0) → c1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
append(Cons(z0, z1), z2) → Cons(z0, append(z1, z2))
append(Nil, z0) → z0
goal(z0, z1) → append(z0, z1)
Tuples:
APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))
S tuples:
APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))
K tuples:none
Defined Rule Symbols:
append, goal
Defined Pair Symbols:
APPEND
Compound Symbols:
c
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
append(Cons(z0, z1), z2) → Cons(z0, append(z1, z2))
append(Nil, z0) → z0
goal(z0, z1) → append(z0, z1)
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))
S tuples:
APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
APPEND
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.
APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))
We considered the (Usable) Rules:none
And the Tuples:
APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(APPEND(x1, x2)) = x1
POL(Cons(x1, x2)) = [1] + x2
POL(c(x1)) = x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))
S tuples:none
K tuples:
APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))
Defined Rule Symbols:none
Defined Pair Symbols:
APPEND
Compound Symbols:
c
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)