(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:
minus(minus(x)) → x
minus(h(x)) → h(minus(x))
minus(f(x, y)) → f(minus(y), minus(x))
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:
minus(minus(x)) → x
(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:
minus(f(x, y)) → f(minus(y), minus(x))
minus(h(x)) → h(minus(x))
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
h0(0) → 0
minus0(0) → 1
minus1(0) → 2
minus1(0) → 3
f1(2, 3) → 1
minus1(0) → 4
h1(4) → 1
f1(2, 3) → 2
f1(2, 3) → 3
f1(2, 3) → 4
h1(4) → 2
h1(4) → 3
h1(4) → 4
(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:
minus(f(z0, z1)) → f(minus(z1), minus(z0))
minus(h(z0)) → h(minus(z0))
Tuples:
MINUS(f(z0, z1)) → c(MINUS(z1), MINUS(z0))
MINUS(h(z0)) → c1(MINUS(z0))
S tuples:
MINUS(f(z0, z1)) → c(MINUS(z1), MINUS(z0))
MINUS(h(z0)) → c1(MINUS(z0))
K tuples:none
Defined Rule Symbols:
minus
Defined Pair Symbols:
MINUS
Compound Symbols:
c, c1
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
minus(f(z0, z1)) → f(minus(z1), minus(z0))
minus(h(z0)) → h(minus(z0))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
MINUS(f(z0, z1)) → c(MINUS(z1), MINUS(z0))
MINUS(h(z0)) → c1(MINUS(z0))
S tuples:
MINUS(f(z0, z1)) → c(MINUS(z1), MINUS(z0))
MINUS(h(z0)) → c1(MINUS(z0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
MINUS
Compound Symbols:
c, c1
(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.
MINUS(h(z0)) → c1(MINUS(z0))
We considered the (Usable) Rules:none
And the Tuples:
MINUS(f(z0, z1)) → c(MINUS(z1), MINUS(z0))
MINUS(h(z0)) → c1(MINUS(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(MINUS(x1)) = x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(f(x1, x2)) = x1 + x2
POL(h(x1)) = [1] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
MINUS(f(z0, z1)) → c(MINUS(z1), MINUS(z0))
MINUS(h(z0)) → c1(MINUS(z0))
S tuples:
MINUS(f(z0, z1)) → c(MINUS(z1), MINUS(z0))
K tuples:
MINUS(h(z0)) → c1(MINUS(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
MINUS
Compound Symbols:
c, c1
(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
MINUS(f(z0, z1)) → c(MINUS(z1), MINUS(z0))
We considered the (Usable) Rules:none
And the Tuples:
MINUS(f(z0, z1)) → c(MINUS(z1), MINUS(z0))
MINUS(h(z0)) → c1(MINUS(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(MINUS(x1)) = x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(f(x1, x2)) = [1] + x1 + x2
POL(h(x1)) = x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
MINUS(f(z0, z1)) → c(MINUS(z1), MINUS(z0))
MINUS(h(z0)) → c1(MINUS(z0))
S tuples:none
K tuples:
MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c(MINUS(z1), MINUS(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
MINUS
Compound Symbols:
c, c1
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(14) BOUNDS(1, 1)