(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:
add0(x', Cons(x, xs)) → add0(Cons(Cons(Nil, Nil), x'), xs)
notEmpty(Cons(x, xs)) → True
notEmpty(Nil) → False
add0(x, Nil) → x
goal(x, y) → add0(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, 3]
transitions:
Cons0(0, 0) → 0
Nil0() → 0
True0() → 0
False0() → 0
add00(0, 0) → 1
notEmpty0(0) → 2
goal0(0, 0) → 3
Nil1() → 6
Nil1() → 7
Cons1(6, 7) → 5
Cons1(5, 0) → 4
add01(4, 0) → 1
True1() → 2
False1() → 2
add01(0, 0) → 3
Cons1(5, 4) → 4
add01(4, 0) → 3
0 → 1
0 → 3
4 → 1
4 → 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:
add0(z0, Cons(z1, z2)) → add0(Cons(Cons(Nil, Nil), z0), z2)
add0(z0, Nil) → z0
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0, z1) → add0(z0, z1)
Tuples:
ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
ADD0(z0, Nil) → c1
NOTEMPTY(Cons(z0, z1)) → c2
NOTEMPTY(Nil) → c3
GOAL(z0, z1) → c4(ADD0(z0, z1))
S tuples:
ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
ADD0(z0, Nil) → c1
NOTEMPTY(Cons(z0, z1)) → c2
NOTEMPTY(Nil) → c3
GOAL(z0, z1) → c4(ADD0(z0, z1))
K tuples:none
Defined Rule Symbols:
add0, notEmpty, goal
Defined Pair Symbols:
ADD0, NOTEMPTY, GOAL
Compound Symbols:
c, c1, c2, c3, c4
(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
GOAL(z0, z1) → c4(ADD0(z0, z1))
Removed 3 trailing nodes:
NOTEMPTY(Cons(z0, z1)) → c2
NOTEMPTY(Nil) → c3
ADD0(z0, Nil) → c1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
add0(z0, Cons(z1, z2)) → add0(Cons(Cons(Nil, Nil), z0), z2)
add0(z0, Nil) → z0
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0, z1) → add0(z0, z1)
Tuples:
ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
S tuples:
ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
K tuples:none
Defined Rule Symbols:
add0, notEmpty, goal
Defined Pair Symbols:
ADD0
Compound Symbols:
c
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
add0(z0, Cons(z1, z2)) → add0(Cons(Cons(Nil, Nil), z0), z2)
add0(z0, Nil) → z0
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0, z1) → add0(z0, z1)
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
S tuples:
ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
ADD0
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.
ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
We considered the (Usable) Rules:none
And the Tuples:
ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(ADD0(x1, x2)) = x1 + [3]x2
POL(Cons(x1, x2)) = [3] + x2
POL(Nil) = 0
POL(c(x1)) = x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
S tuples:none
K tuples:
ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
Defined Rule Symbols:none
Defined Pair Symbols:
ADD0
Compound Symbols:
c
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)