(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
times(z0, s(z1)) → plus(z0, times(z1, z0))
Tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
S tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
K tuples:none
Defined Rule Symbols:
plus, times
Defined Pair Symbols:
PLUS, TIMES
Compound Symbols:
c, c1
(3) CdtRuleRemovalProof (UPPER BOUND(ADD(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(z0, times(z1, z0)), TIMES(z1, z0))
We considered the (Usable) Rules:none
And the Tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(PLUS(x1, x2)) = 0
POL(TIMES(x1, x2)) = x1 + x2
POL(c(x1, x2)) = x1 + x2
POL(c1(x1, x2)) = x1 + x2
POL(plus(x1, x2)) = [3]
POL(s(x1)) = [1] + x1
POL(times(x1, x2)) = [3]
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
times(z0, s(z1)) → plus(z0, times(z1, z0))
Tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
S tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
K tuples:
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
Defined Rule Symbols:
plus, times
Defined Pair Symbols:
PLUS, TIMES
Compound Symbols:
c, c1
(5) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
We considered the (Usable) Rules:none
And the Tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(PLUS(x1, x2)) = x1
POL(TIMES(x1, x2)) = [2]x1·x2
POL(c(x1, x2)) = x1 + x2
POL(c1(x1, x2)) = x1 + x2
POL(plus(x1, x2)) = [2] + [2]x1 + x2
POL(s(x1)) = [3] + x1
POL(times(x1, x2)) = 0
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
times(z0, s(z1)) → plus(z0, times(z1, z0))
Tuples:
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
S tuples:none
K tuples:
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
Defined Rule Symbols:
plus, times
Defined Pair Symbols:
PLUS, TIMES
Compound Symbols:
c, c1
(7) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(8) BOUNDS(1, 1)