(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, s(y)) → if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) → s(quot(minus(x, y), y))
if_quot(false, x, y) → 0
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(z0, s(z1)) → if_quot(le(s(z1), z0), z0, s(z1))
if_quot(true, z0, z1) → s(quot(minus(z0, z1), z1))
if_quot(false, z0, z1) → 0
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(z0, s(z1)) → c5(IF_QUOT(le(s(z1), z0), z0, s(z1)), LE(s(z1), z0))
IF_QUOT(true, z0, z1) → c6(QUOT(minus(z0, z1), z1), MINUS(z0, z1))
S tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(z0, s(z1)) → c5(IF_QUOT(le(s(z1), z0), z0, s(z1)), LE(s(z1), z0))
IF_QUOT(true, z0, z1) → c6(QUOT(minus(z0, z1), z1), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
minus, le, quot, if_quot
Defined Pair Symbols:
MINUS, LE, QUOT, IF_QUOT
Compound Symbols:
c1, c4, c5, c6
(3) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
QUOT(
z0,
s(
z1)) →
c5(
IF_QUOT(
le(
s(
z1),
z0),
z0,
s(
z1)),
LE(
s(
z1),
z0)) by
QUOT(0, s(z0)) → c5(IF_QUOT(false, 0, s(z0)), LE(s(z0), 0))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
QUOT(x0, s(x1)) → c5
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(z0, s(z1)) → if_quot(le(s(z1), z0), z0, s(z1))
if_quot(true, z0, z1) → s(quot(minus(z0, z1), z1))
if_quot(false, z0, z1) → 0
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
IF_QUOT(true, z0, z1) → c6(QUOT(minus(z0, z1), z1), MINUS(z0, z1))
QUOT(0, s(z0)) → c5(IF_QUOT(false, 0, s(z0)), LE(s(z0), 0))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
QUOT(x0, s(x1)) → c5
S tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
IF_QUOT(true, z0, z1) → c6(QUOT(minus(z0, z1), z1), MINUS(z0, z1))
QUOT(0, s(z0)) → c5(IF_QUOT(false, 0, s(z0)), LE(s(z0), 0))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
QUOT(x0, s(x1)) → c5
K tuples:none
Defined Rule Symbols:
minus, le, quot, if_quot
Defined Pair Symbols:
MINUS, LE, IF_QUOT, QUOT
Compound Symbols:
c1, c4, c6, c5, c5
(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
QUOT(0, s(z0)) → c5(IF_QUOT(false, 0, s(z0)), LE(s(z0), 0))
QUOT(x0, s(x1)) → c5
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(z0, s(z1)) → if_quot(le(s(z1), z0), z0, s(z1))
if_quot(true, z0, z1) → s(quot(minus(z0, z1), z1))
if_quot(false, z0, z1) → 0
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
IF_QUOT(true, z0, z1) → c6(QUOT(minus(z0, z1), z1), MINUS(z0, z1))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
S tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
IF_QUOT(true, z0, z1) → c6(QUOT(minus(z0, z1), z1), MINUS(z0, z1))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
K tuples:none
Defined Rule Symbols:
minus, le, quot, if_quot
Defined Pair Symbols:
MINUS, LE, IF_QUOT, QUOT
Compound Symbols:
c1, c4, c6, c5
(7) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
IF_QUOT(
true,
z0,
z1) →
c6(
QUOT(
minus(
z0,
z1),
z1),
MINUS(
z0,
z1)) by
IF_QUOT(true, z0, 0) → c6(QUOT(z0, 0), MINUS(z0, 0))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF_QUOT(true, x0, x1) → c6
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(z0, s(z1)) → if_quot(le(s(z1), z0), z0, s(z1))
if_quot(true, z0, z1) → s(quot(minus(z0, z1), z1))
if_quot(false, z0, z1) → 0
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, z0, 0) → c6(QUOT(z0, 0), MINUS(z0, 0))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF_QUOT(true, x0, x1) → c6
S tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, z0, 0) → c6(QUOT(z0, 0), MINUS(z0, 0))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF_QUOT(true, x0, x1) → c6
K tuples:none
Defined Rule Symbols:
minus, le, quot, if_quot
Defined Pair Symbols:
MINUS, LE, QUOT, IF_QUOT
Compound Symbols:
c1, c4, c5, c6, c6
(9) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
IF_QUOT(true, z0, 0) → c6(QUOT(z0, 0), MINUS(z0, 0))
IF_QUOT(true, x0, x1) → c6
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(z0, s(z1)) → if_quot(le(s(z1), z0), z0, s(z1))
if_quot(true, z0, z1) → s(quot(minus(z0, z1), z1))
if_quot(false, z0, z1) → 0
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
S tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
K tuples:none
Defined Rule Symbols:
minus, le, quot, if_quot
Defined Pair Symbols:
MINUS, LE, QUOT, IF_QUOT
Compound Symbols:
c1, c4, c5, c6
(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
We considered the (Usable) Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
And the Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [2]
POL(IF_QUOT(x1, x2, x3)) = [1] + [2]x2
POL(LE(x1, x2)) = [2]
POL(MINUS(x1, x2)) = [1]
POL(QUOT(x1, x2)) = [4] + [2]x1
POL(c1(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1, x2)) = x1 + x2
POL(c6(x1, x2)) = x1 + x2
POL(false) = [3]
POL(le(x1, x2)) = x1 + [3]x2
POL(minus(x1, x2)) = x1
POL(s(x1)) = [4] + x1
POL(true) = 0
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(z0, s(z1)) → if_quot(le(s(z1), z0), z0, s(z1))
if_quot(true, z0, z1) → s(quot(minus(z0, z1), z1))
if_quot(false, z0, z1) → 0
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
S tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
K tuples:
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
Defined Rule Symbols:
minus, le, quot, if_quot
Defined Pair Symbols:
MINUS, LE, QUOT, IF_QUOT
Compound Symbols:
c1, c4, c5, c6
(13) CdtPolyRedPairProof (UPPER BOUND (ADD(O(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(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
And the Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(IF_QUOT(x1, x2, x3)) = [2]x2 + x22
POL(LE(x1, x2)) = 0
POL(MINUS(x1, x2)) = [1] + [2]x1
POL(QUOT(x1, x2)) = [2]x1 + x12
POL(c1(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1, x2)) = x1 + x2
POL(c6(x1, x2)) = x1 + x2
POL(false) = 0
POL(le(x1, x2)) = 0
POL(minus(x1, x2)) = x1
POL(s(x1)) = [1] + x1
POL(true) = 0
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(z0, s(z1)) → if_quot(le(s(z1), z0), z0, s(z1))
if_quot(true, z0, z1) → s(quot(minus(z0, z1), z1))
if_quot(false, z0, z1) → 0
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
S tuples:
LE(s(z0), s(z1)) → c4(LE(z0, z1))
K tuples:
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
Defined Rule Symbols:
minus, le, quot, if_quot
Defined Pair Symbols:
MINUS, LE, QUOT, IF_QUOT
Compound Symbols:
c1, c4, c5, c6
(15) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
LE(s(z0), s(z1)) → c4(LE(z0, z1))
We considered the (Usable) Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
And the Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(IF_QUOT(x1, x2, x3)) = [2]x22
POL(LE(x1, x2)) = x2
POL(MINUS(x1, x2)) = 0
POL(QUOT(x1, x2)) = x1 + [2]x12
POL(c1(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1, x2)) = x1 + x2
POL(c6(x1, x2)) = x1 + x2
POL(false) = 0
POL(le(x1, x2)) = 0
POL(minus(x1, x2)) = x1
POL(s(x1)) = [1] + x1
POL(true) = 0
(16) Obligation:
Complexity Dependency Tuples Problem
Rules:
minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
quot(z0, s(z1)) → if_quot(le(s(z1), z0), z0, s(z1))
if_quot(true, z0, z1) → s(quot(minus(z0, z1), z1))
if_quot(false, z0, z1) → 0
Tuples:
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
S tuples:none
K tuples:
QUOT(s(z1), s(z0)) → c5(IF_QUOT(le(z0, z1), s(z1), s(z0)), LE(s(z0), s(z1)))
IF_QUOT(true, s(z0), s(z1)) → c6(QUOT(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
LE(s(z0), s(z1)) → c4(LE(z0, z1))
Defined Rule Symbols:
minus, le, quot, if_quot
Defined Pair Symbols:
MINUS, LE, QUOT, IF_QUOT
Compound Symbols:
c1, c4, c5, c6
(17) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(18) BOUNDS(O(1), O(1))