(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)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(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)) → minus(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:
MINUS(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(0, s(z0)) → c2
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
MINUS(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(0, s(z0)) → c2
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
minus, quot
Defined Pair Symbols:
MINUS, QUOT
Compound Symbols:
c, c1, c2, c3
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
MINUS(z0, 0) → c
QUOT(0, s(z0)) → c2
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
minus, quot
Defined Pair Symbols:
MINUS, QUOT
Compound Symbols:
c1, c3
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
minus
Defined Pair Symbols:
MINUS, QUOT
Compound Symbols:
c1, c3
(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
And the Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(MINUS(x1, x2)) = 0
POL(QUOT(x1, x2)) = x1
POL(c1(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = x1
POL(s(x1)) = [1] + x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:
minus
Defined Pair Symbols:
MINUS, QUOT
Compound Symbols:
c1, c3
(9) 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:
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
And the Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(MINUS(x1, x2)) = [2]x1
POL(QUOT(x1, x2)) = [2]x12
POL(c1(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = x1
POL(s(x1)) = [1] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:none
K tuples:
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
Defined Rule Symbols:
minus
Defined Pair Symbols:
MINUS, QUOT
Compound Symbols:
c1, c3
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)