(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxRelTRS could be proven to be
BOUNDS(1, n^1).
The TRS R consists of the following rules:
map(Cons(x, xs)) → Cons(f(x), map(xs))
map(Nil) → Nil
goal(xs) → map(xs)
f(x) → *(x, x)
+Full(S(x), y) → +Full(x, S(y))
+Full(0, y) → y
The (relative) TRS S consists of the following rules:
*(x, S(S(y))) → +(x, *(x, S(y)))
*(x, S(0)) → x
*(x, 0) → 0
*(0, y) → 0
Rewrite Strategy: INNERMOST
(1) RelTrsToTrsProof (UPPER BOUND(ID) transformation)
transformed relative TRS to TRS
(2) 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:
map(Cons(x, xs)) → Cons(f(x), map(xs))
map(Nil) → Nil
goal(xs) → map(xs)
f(x) → *(x, x)
+Full(S(x), y) → +Full(x, S(y))
+Full(0, y) → y
*(x, S(S(y))) → +(x, *(x, S(y)))
*(x, S(0)) → x
*(x, 0) → 0
*(0, y) → 0
Rewrite Strategy: INNERMOST
(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
map(Cons(z0, z1)) → Cons(f(z0), map(z1))
map(Nil) → Nil
goal(z0) → map(z0)
f(z0) → *(z0, z0)
+Full(S(z0), z1) → +Full(z0, S(z1))
+Full(0, z0) → z0
*(z0, S(S(z1))) → +(z0, *(z0, S(z1)))
*(z0, S(0)) → z0
*(z0, 0) → 0
*(0, z0) → 0
Tuples:
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
MAP(Nil) → c1
GOAL(z0) → c2(MAP(z0))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
+FULL(0, z0) → c5
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
*'(z0, S(0)) → c7
*'(z0, 0) → c8
*'(0, z0) → c9
S tuples:
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
MAP(Nil) → c1
GOAL(z0) → c2(MAP(z0))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
+FULL(0, z0) → c5
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
*'(z0, S(0)) → c7
*'(z0, 0) → c8
*'(0, z0) → c9
K tuples:none
Defined Rule Symbols:
map, goal, f, +Full, *
Defined Pair Symbols:
MAP, GOAL, F, +FULL, *'
Compound Symbols:
c, c1, c2, c3, c4, c5, c6, c7, c8, c9
(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
GOAL(z0) → c2(MAP(z0))
Removed 5 trailing nodes:
*'(0, z0) → c9
*'(z0, 0) → c8
*'(z0, S(0)) → c7
MAP(Nil) → c1
+FULL(0, z0) → c5
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
map(Cons(z0, z1)) → Cons(f(z0), map(z1))
map(Nil) → Nil
goal(z0) → map(z0)
f(z0) → *(z0, z0)
+Full(S(z0), z1) → +Full(z0, S(z1))
+Full(0, z0) → z0
*(z0, S(S(z1))) → +(z0, *(z0, S(z1)))
*(z0, S(0)) → z0
*(z0, 0) → 0
*(0, z0) → 0
Tuples:
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
S tuples:
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
K tuples:none
Defined Rule Symbols:
map, goal, f, +Full, *
Defined Pair Symbols:
MAP, F, +FULL, *'
Compound Symbols:
c, c3, c4, c6
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
map(Cons(z0, z1)) → Cons(f(z0), map(z1))
map(Nil) → Nil
goal(z0) → map(z0)
f(z0) → *(z0, z0)
+Full(S(z0), z1) → +Full(z0, S(z1))
+Full(0, z0) → z0
*(z0, S(S(z1))) → +(z0, *(z0, S(z1)))
*(z0, S(0)) → z0
*(z0, 0) → 0
*(0, z0) → 0
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
S tuples:
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
MAP, F, +FULL, *'
Compound Symbols:
c, c3, c4, c6
(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.
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
We considered the (Usable) Rules:none
And the Tuples:
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(*'(x1, x2)) = 0
POL(+FULL(x1, x2)) = [3]x1 + [2]x2
POL(Cons(x1, x2)) = [2] + x2
POL(F(x1)) = [3]
POL(MAP(x1)) = [2]x1
POL(S(x1)) = [3] + x1
POL(c(x1, x2)) = x1 + x2
POL(c3(x1)) = x1
POL(c4(x1)) = x1
POL(c6(x1)) = x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
S tuples:
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
K tuples:
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
Defined Rule Symbols:none
Defined Pair Symbols:
MAP, F, +FULL, *'
Compound Symbols:
c, c3, c4, c6
(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
We considered the (Usable) Rules:none
And the Tuples:
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(*'(x1, x2)) = x2
POL(+FULL(x1, x2)) = x1
POL(Cons(x1, x2)) = x1 + x2
POL(F(x1)) = x1
POL(MAP(x1)) = x1
POL(S(x1)) = [1] + x1
POL(c(x1, x2)) = x1 + x2
POL(c3(x1)) = x1
POL(c4(x1)) = x1
POL(c6(x1)) = x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
S tuples:none
K tuples:
MAP(Cons(z0, z1)) → c(F(z0), MAP(z1))
F(z0) → c3(*'(z0, z0))
+FULL(S(z0), z1) → c4(+FULL(z0, S(z1)))
*'(z0, S(S(z1))) → c6(*'(z0, S(z1)))
Defined Rule Symbols:none
Defined Pair Symbols:
MAP, F, +FULL, *'
Compound Symbols:
c, c3, c4, c6
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(14) BOUNDS(1, 1)