(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:
is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, l1)
ifappend(l1, l2, nil) → l2
ifappend(l1, l2, cons(x, l)) → cons(x, append(l, l2))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
is_empty(nil) → true
is_empty(cons(z0, z1)) → false
hd(cons(z0, z1)) → z0
tl(cons(z0, z1)) → z1
append(z0, z1) → ifappend(z0, z1, z0)
ifappend(z0, z1, nil) → z1
ifappend(z0, z1, cons(z2, z3)) → cons(z2, append(z3, z1))
Tuples:
IS_EMPTY(nil) → c
IS_EMPTY(cons(z0, z1)) → c1
HD(cons(z0, z1)) → c2
TL(cons(z0, z1)) → c3
APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0))
IFAPPEND(z0, z1, nil) → c5
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
S tuples:
IS_EMPTY(nil) → c
IS_EMPTY(cons(z0, z1)) → c1
HD(cons(z0, z1)) → c2
TL(cons(z0, z1)) → c3
APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0))
IFAPPEND(z0, z1, nil) → c5
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
K tuples:none
Defined Rule Symbols:
is_empty, hd, tl, append, ifappend
Defined Pair Symbols:
IS_EMPTY, HD, TL, APPEND, IFAPPEND
Compound Symbols:
c, c1, c2, c3, c4, c5, c6
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 5 trailing nodes:
HD(cons(z0, z1)) → c2
TL(cons(z0, z1)) → c3
IFAPPEND(z0, z1, nil) → c5
IS_EMPTY(cons(z0, z1)) → c1
IS_EMPTY(nil) → c
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
is_empty(nil) → true
is_empty(cons(z0, z1)) → false
hd(cons(z0, z1)) → z0
tl(cons(z0, z1)) → z1
append(z0, z1) → ifappend(z0, z1, z0)
ifappend(z0, z1, nil) → z1
ifappend(z0, z1, cons(z2, z3)) → cons(z2, append(z3, z1))
Tuples:
APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0))
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
S tuples:
APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0))
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
K tuples:none
Defined Rule Symbols:
is_empty, hd, tl, append, ifappend
Defined Pair Symbols:
APPEND, IFAPPEND
Compound Symbols:
c4, c6
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
is_empty(nil) → true
is_empty(cons(z0, z1)) → false
hd(cons(z0, z1)) → z0
tl(cons(z0, z1)) → z1
append(z0, z1) → ifappend(z0, z1, z0)
ifappend(z0, z1, nil) → z1
ifappend(z0, z1, cons(z2, z3)) → cons(z2, append(z3, z1))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0))
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
S tuples:
APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0))
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
APPEND, IFAPPEND
Compound Symbols:
c4, c6
(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.
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
We considered the (Usable) Rules:none
And the Tuples:
APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0))
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(APPEND(x1, x2)) = x1
POL(IFAPPEND(x1, x2, x3)) = x3
POL(c4(x1)) = x1
POL(c6(x1)) = x1
POL(cons(x1, x2)) = [1] + x2
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0))
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
S tuples:
APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0))
K tuples:
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
Defined Rule Symbols:none
Defined Pair Symbols:
APPEND, IFAPPEND
Compound Symbols:
c4, c6
(9) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
APPEND(z0, z1) → c4(IFAPPEND(z0, z1, z0))
IFAPPEND(z0, z1, cons(z2, z3)) → c6(APPEND(z3, z1))
Now S is empty
(10) BOUNDS(1, 1)