(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(0, z0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(s(z0), z1) → s(plus(z0, z1))
Tuples:
TIMES(z0, 0) → c
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, 0) → c2
PLUS(0, z0) → c3
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
S tuples:
TIMES(z0, 0) → c
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, 0) → c2
PLUS(0, z0) → c3
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
K tuples:none
Defined Rule Symbols:
times, plus
Defined Pair Symbols:
TIMES, PLUS
Compound Symbols:
c, c1, c2, c3, c4, c5
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 3 trailing nodes:
TIMES(z0, 0) → c
PLUS(z0, 0) → c2
PLUS(0, z0) → c3
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(0, z0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(s(z0), z1) → s(plus(z0, z1))
Tuples:
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
S tuples:
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
K tuples:none
Defined Rule Symbols:
times, plus
Defined Pair Symbols:
TIMES, PLUS
Compound Symbols:
c1, c4, c5
(5) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [3]
POL(PLUS(x1, x2)) = [3]
POL(TIMES(x1, x2)) = [4]x2
POL(c1(x1, x2)) = x1 + x2
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(plus(x1, x2)) = [3]x2
POL(s(x1)) = [1] + x1
POL(times(x1, x2)) = [3]x1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(0, z0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(s(z0), z1) → s(plus(z0, z1))
Tuples:
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
S tuples:
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
K tuples:
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
Defined Rule Symbols:
times, plus
Defined Pair Symbols:
TIMES, PLUS
Compound Symbols:
c1, c4, c5
(7) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^2))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(PLUS(x1, x2)) = [1] + x2
POL(TIMES(x1, x2)) = [2]x2 + [2]x1·x2
POL(c1(x1, x2)) = x1 + x2
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(plus(x1, x2)) = 0
POL(s(x1)) = [2] + x1
POL(times(x1, x2)) = 0
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(0, z0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(s(z0), z1) → s(plus(z0, z1))
Tuples:
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
S tuples:
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
K tuples:
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
Defined Rule Symbols:
times, plus
Defined Pair Symbols:
TIMES, PLUS
Compound Symbols:
c1, c4, c5
(9) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^3))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
We considered the (Usable) Rules:
plus(z0, s(z1)) → s(plus(z0, z1))
plus(0, z0) → z0
plus(z0, 0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(z0, s(z1)) → plus(times(z0, z1), z0)
times(z0, 0) → 0
And the Tuples:
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(PLUS(x1, x2)) = x1 + x2 + x22
POL(TIMES(x1, x2)) = x12·x2 + x1·x22
POL(c1(x1, x2)) = x1 + x2
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(plus(x1, x2)) = x1 + x2
POL(s(x1)) = [1] + x1
POL(times(x1, x2)) = x1·x2
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(0, z0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(s(z0), z1) → s(plus(z0, z1))
Tuples:
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
S tuples:none
K tuples:
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
Defined Rule Symbols:
times, plus
Defined Pair Symbols:
TIMES, PLUS
Compound Symbols:
c1, c4, c5
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(O(1), O(1))