(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:
quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(quot(x, s(z), s(z)))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
quot(0, s(z0), s(z1)) → 0
quot(s(z0), s(z1), z2) → quot(z0, z1, z2)
quot(z0, 0, s(z1)) → s(quot(z0, s(z1), s(z1)))
Tuples:
QUOT(0, s(z0), s(z1)) → c
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))
S tuples:
QUOT(0, s(z0), s(z1)) → c
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:
quot
Defined Pair Symbols:
QUOT
Compound Symbols:
c, c1, c2
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
QUOT(0, s(z0), s(z1)) → c
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
quot(0, s(z0), s(z1)) → 0
quot(s(z0), s(z1), z2) → quot(z0, z1, z2)
quot(z0, 0, s(z1)) → s(quot(z0, s(z1), s(z1)))
Tuples:
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))
S tuples:
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:
quot
Defined Pair Symbols:
QUOT
Compound Symbols:
c1, c2
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
quot(0, s(z0), s(z1)) → 0
quot(s(z0), s(z1), z2) → quot(z0, z1, z2)
quot(z0, 0, s(z1)) → s(quot(z0, s(z1), s(z1)))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))
S tuples:
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
QUOT
Compound Symbols:
c1, c2
(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), z2) → c1(QUOT(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(QUOT(x1, x2, x3)) = x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(s(x1)) = [1] + x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))
S tuples:
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))
K tuples:
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
Defined Rule Symbols:none
Defined Pair Symbols:
QUOT
Compound Symbols:
c1, c2
(9) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
QUOT(z0, 0, s(z1)) → c2(QUOT(z0, s(z1), s(z1)))
QUOT(s(z0), s(z1), z2) → c1(QUOT(z0, z1, z2))
Now S is empty
(10) BOUNDS(1, 1)