(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:
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(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:
pred(s(z0)) → z0
minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:
PRED(s(z0)) → c
MINUS(z0, 0) → c1
MINUS(z0, s(z1)) → c2(PRED(minus(z0, z1)), MINUS(z0, z1))
QUOT(0, s(z0)) → c3
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
PRED(s(z0)) → c
MINUS(z0, 0) → c1
MINUS(z0, s(z1)) → c2(PRED(minus(z0, z1)), MINUS(z0, z1))
QUOT(0, s(z0)) → c3
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
pred, minus, quot
Defined Pair Symbols:
PRED, MINUS, QUOT
Compound Symbols:
c, c1, c2, c3, c4
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 3 trailing nodes:
QUOT(0, s(z0)) → c3
MINUS(z0, 0) → c1
PRED(s(z0)) → c
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
pred(s(z0)) → z0
minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:
MINUS(z0, s(z1)) → c2(PRED(minus(z0, z1)), MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
MINUS(z0, s(z1)) → c2(PRED(minus(z0, z1)), MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
pred, minus, quot
Defined Pair Symbols:
MINUS, QUOT
Compound Symbols:
c2, c4
(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
pred(s(z0)) → z0
minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
pred, minus, quot
Defined Pair Symbols:
QUOT, MINUS
Compound Symbols:
c4, c2
(7) 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)))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
pred(s(z0)) → z0
Tuples:
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
minus, pred
Defined Pair Symbols:
QUOT, MINUS
Compound Symbols:
c4, c2
(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.
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:
minus(z0, s(z1)) → pred(minus(z0, z1))
minus(z0, 0) → z0
pred(s(z0)) → z0
And the Tuples:
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(z0, s(z1)) → c2(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(c2(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = x1
POL(pred(x1)) = x1
POL(s(x1)) = [2] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
pred(s(z0)) → z0
Tuples:
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
K tuples:
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:
minus, pred
Defined Pair Symbols:
QUOT, MINUS
Compound Symbols:
c4, c2
(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(z0, s(z1)) → c2(MINUS(z0, z1))
We considered the (Usable) Rules:
minus(z0, s(z1)) → pred(minus(z0, z1))
minus(z0, 0) → z0
pred(s(z0)) → z0
And the Tuples:
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(MINUS(x1, x2)) = [1] + [2]x1 + x2
POL(QUOT(x1, x2)) = [2]x1·x2 + [2]x12
POL(c2(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(minus(x1, x2)) = x1
POL(pred(x1)) = x1
POL(s(x1)) = [1] + x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(z0, s(z1)) → pred(minus(z0, z1))
pred(s(z0)) → z0
Tuples:
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
S tuples:none
K tuples:
QUOT(s(z0), s(z1)) → c4(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(z0, s(z1)) → c2(MINUS(z0, z1))
Defined Rule Symbols:
minus, pred
Defined Pair Symbols:
QUOT, MINUS
Compound Symbols:
c4, c2
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(14) BOUNDS(1, 1)