(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, n^2).
The TRS R consists of the following rules:
minus(X, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → p(minus(z0, z1))
p(s(z0)) → z0
div(0, s(z0)) → 0
div(s(z0), s(z1)) → s(div(minus(z0, z1), s(z1)))
Tuples:
MINUS(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(P(minus(z0, z1)), MINUS(z0, z1))
P(s(z0)) → c2
DIV(0, s(z0)) → c3
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
MINUS(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(P(minus(z0, z1)), MINUS(z0, z1))
P(s(z0)) → c2
DIV(0, s(z0)) → c3
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
minus, p, div
Defined Pair Symbols:
MINUS, P, DIV
Compound Symbols:
c, c1, c2, c3, c4
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 3 trailing nodes:
P(s(z0)) → c2
DIV(0, s(z0)) → c3
MINUS(z0, 0) → c
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → p(minus(z0, z1))
p(s(z0)) → z0
div(0, s(z0)) → 0
div(s(z0), s(z1)) → s(div(minus(z0, z1), s(z1)))
Tuples:
MINUS(s(z0), s(z1)) → c1(P(minus(z0, z1)), MINUS(z0, z1))
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
MINUS(s(z0), s(z1)) → c1(P(minus(z0, z1)), MINUS(z0, z1))
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
minus, p, div
Defined Pair Symbols:
MINUS, DIV
Compound Symbols:
c1, c4
(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → p(minus(z0, z1))
p(s(z0)) → z0
div(0, s(z0)) → 0
div(s(z0), s(z1)) → s(div(minus(z0, z1), s(z1)))
Tuples:
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
minus, p, div
Defined Pair Symbols:
DIV, MINUS
Compound Symbols:
c4, c1
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
div(0, s(z0)) → 0
div(s(z0), s(z1)) → s(div(minus(z0, z1), s(z1)))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → p(minus(z0, z1))
p(s(z0)) → z0
Tuples:
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
minus, p
Defined Pair Symbols:
DIV, MINUS
Compound Symbols:
c4, 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.
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:
p(s(z0)) → z0
minus(z0, 0) → z0
minus(s(z0), s(z1)) → p(minus(z0, z1))
And the Tuples:
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(DIV(x1, x2)) = x1
POL(MINUS(x1, x2)) = 0
POL(c1(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = x1
POL(p(x1)) = x1
POL(s(x1)) = [1] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → p(minus(z0, z1))
p(s(z0)) → z0
Tuples:
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:
minus, p
Defined Pair Symbols:
DIV, MINUS
Compound Symbols:
c4, c1
(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
We considered the (Usable) Rules:
p(s(z0)) → z0
minus(z0, 0) → z0
minus(s(z0), s(z1)) → p(minus(z0, z1))
And the Tuples:
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(DIV(x1, x2)) = x12
POL(MINUS(x1, x2)) = [2]x1
POL(c1(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = x1
POL(p(x1)) = [1] + x1
POL(s(x1)) = [1] + x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → p(minus(z0, z1))
p(s(z0)) → z0
Tuples:
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:none
K tuples:
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
Defined Rule Symbols:
minus, p
Defined Pair Symbols:
DIV, MINUS
Compound Symbols:
c4, c1
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(14) BOUNDS(1, 1)