YES(O(1),O(n^1)) 0.00/0.32 YES(O(1),O(n^1)) 0.00/0.32 0.00/0.32 We are left with following problem, upon which TcT provides the 0.00/0.32 certificate YES(O(1),O(n^1)). 0.00/0.32 0.00/0.32 Strict Trs: 0.00/0.32 { is_empty(nil()) -> true() 0.00/0.32 , is_empty(cons(x, l)) -> false() 0.00/0.32 , hd(cons(x, l)) -> x 0.00/0.32 , tl(cons(x, l)) -> l 0.00/0.32 , append(l1, l2) -> ifappend(l1, l2, l1) 0.00/0.32 , ifappend(l1, l2, nil()) -> l2 0.00/0.32 , ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2)) } 0.00/0.32 Obligation: 0.00/0.32 runtime complexity 0.00/0.32 Answer: 0.00/0.32 YES(O(1),O(n^1)) 0.00/0.32 0.00/0.32 The weightgap principle applies (using the following nonconstant 0.00/0.32 growth matrix-interpretation) 0.00/0.32 0.00/0.32 The following argument positions are usable: 0.00/0.32 Uargs(cons) = {2} 0.00/0.32 0.00/0.32 TcT has computed the following matrix interpretation satisfying 0.00/0.32 not(EDA) and not(IDA(1)). 0.00/0.32 0.00/0.32 [is_empty](x1) = [1] x1 + [7] 0.00/0.32 0.00/0.32 [nil] = [7] 0.00/0.32 0.00/0.32 [true] = [5] 0.00/0.32 0.00/0.32 [cons](x1, x2) = [1] x1 + [1] x2 + [7] 0.00/0.32 0.00/0.32 [false] = [5] 0.00/0.32 0.00/0.32 [hd](x1) = [1] x1 + [7] 0.00/0.32 0.00/0.32 [tl](x1) = [1] x1 + [7] 0.00/0.32 0.00/0.32 [append](x1, x2) = [1] x1 + [1] x2 + [7] 0.00/0.32 0.00/0.32 [ifappend](x1, x2, x3) = [1] x2 + [1] x3 + [7] 0.00/0.32 0.00/0.32 The order satisfies the following ordering constraints: 0.00/0.32 0.00/0.32 [is_empty(nil())] = [14] 0.00/0.32 > [5] 0.00/0.32 = [true()] 0.00/0.32 0.00/0.32 [is_empty(cons(x, l))] = [1] x + [1] l + [14] 0.00/0.32 > [5] 0.00/0.32 = [false()] 0.00/0.32 0.00/0.32 [hd(cons(x, l))] = [1] x + [1] l + [14] 0.00/0.32 > [1] x + [0] 0.00/0.32 = [x] 0.00/0.32 0.00/0.32 [tl(cons(x, l))] = [1] x + [1] l + [14] 0.00/0.32 > [1] l + [0] 0.00/0.32 = [l] 0.00/0.32 0.00/0.32 [append(l1, l2)] = [1] l1 + [1] l2 + [7] 0.00/0.32 >= [1] l1 + [1] l2 + [7] 0.00/0.32 = [ifappend(l1, l2, l1)] 0.00/0.32 0.00/0.32 [ifappend(l1, l2, nil())] = [1] l2 + [14] 0.00/0.32 > [1] l2 + [0] 0.00/0.32 = [l2] 0.00/0.32 0.00/0.32 [ifappend(l1, l2, cons(x, l))] = [1] x + [1] l + [1] l2 + [14] 0.00/0.32 >= [1] x + [1] l + [1] l2 + [14] 0.00/0.32 = [cons(x, append(l, l2))] 0.00/0.32 0.00/0.32 0.00/0.32 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.32 0.00/0.32 We are left with following problem, upon which TcT provides the 0.00/0.32 certificate YES(O(1),O(n^1)). 0.00/0.32 0.00/0.32 Strict Trs: 0.00/0.32 { append(l1, l2) -> ifappend(l1, l2, l1) 0.00/0.32 , ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2)) } 0.00/0.32 Weak Trs: 0.00/0.32 { is_empty(nil()) -> true() 0.00/0.32 , is_empty(cons(x, l)) -> false() 0.00/0.32 , hd(cons(x, l)) -> x 0.00/0.32 , tl(cons(x, l)) -> l 0.00/0.32 , ifappend(l1, l2, nil()) -> l2 } 0.00/0.32 Obligation: 0.00/0.32 runtime complexity 0.00/0.32 Answer: 0.00/0.32 YES(O(1),O(n^1)) 0.00/0.32 0.00/0.32 The weightgap principle applies (using the following nonconstant 0.00/0.32 growth matrix-interpretation) 0.00/0.32 0.00/0.32 The following argument positions are usable: 0.00/0.32 Uargs(cons) = {2} 0.00/0.32 0.00/0.32 TcT has computed the following matrix interpretation satisfying 0.00/0.32 not(EDA) and not(IDA(1)). 0.00/0.32 0.00/0.32 [is_empty](x1) = [1] x1 + [7] 0.00/0.32 0.00/0.32 [nil] = [7] 0.00/0.32 0.00/0.32 [true] = [5] 0.00/0.32 0.00/0.32 [cons](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.32 0.00/0.32 [false] = [7] 0.00/0.32 0.00/0.32 [hd](x1) = [1] x1 + [7] 0.00/0.32 0.00/0.32 [tl](x1) = [1] x1 + [7] 0.00/0.32 0.00/0.32 [append](x1, x2) = [1] x1 + [1] x2 + [1] 0.00/0.32 0.00/0.32 [ifappend](x1, x2, x3) = [1] x2 + [1] x3 + [0] 0.00/0.32 0.00/0.32 The order satisfies the following ordering constraints: 0.00/0.32 0.00/0.32 [is_empty(nil())] = [14] 0.00/0.32 > [5] 0.00/0.32 = [true()] 0.00/0.32 0.00/0.32 [is_empty(cons(x, l))] = [1] x + [1] l + [7] 0.00/0.32 >= [7] 0.00/0.32 = [false()] 0.00/0.32 0.00/0.32 [hd(cons(x, l))] = [1] x + [1] l + [7] 0.00/0.32 > [1] x + [0] 0.00/0.32 = [x] 0.00/0.32 0.00/0.32 [tl(cons(x, l))] = [1] x + [1] l + [7] 0.00/0.32 > [1] l + [0] 0.00/0.32 = [l] 0.00/0.32 0.00/0.32 [append(l1, l2)] = [1] l1 + [1] l2 + [1] 0.00/0.32 > [1] l1 + [1] l2 + [0] 0.00/0.32 = [ifappend(l1, l2, l1)] 0.00/0.32 0.00/0.32 [ifappend(l1, l2, nil())] = [1] l2 + [7] 0.00/0.32 > [1] l2 + [0] 0.00/0.32 = [l2] 0.00/0.32 0.00/0.32 [ifappend(l1, l2, cons(x, l))] = [1] x + [1] l + [1] l2 + [0] 0.00/0.32 ? [1] x + [1] l + [1] l2 + [1] 0.00/0.32 = [cons(x, append(l, l2))] 0.00/0.32 0.00/0.32 0.00/0.32 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.32 0.00/0.32 We are left with following problem, upon which TcT provides the 0.00/0.32 certificate YES(O(1),O(n^1)). 0.00/0.32 0.00/0.32 Strict Trs: 0.00/0.32 { ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2)) } 0.00/0.32 Weak Trs: 0.00/0.32 { is_empty(nil()) -> true() 0.00/0.32 , is_empty(cons(x, l)) -> false() 0.00/0.32 , hd(cons(x, l)) -> x 0.00/0.32 , tl(cons(x, l)) -> l 0.00/0.32 , append(l1, l2) -> ifappend(l1, l2, l1) 0.00/0.32 , ifappend(l1, l2, nil()) -> l2 } 0.00/0.32 Obligation: 0.00/0.32 runtime complexity 0.00/0.32 Answer: 0.00/0.32 YES(O(1),O(n^1)) 0.00/0.32 0.00/0.32 We use the processor 'matrix interpretation of dimension 1' to 0.00/0.32 orient following rules strictly. 0.00/0.32 0.00/0.32 Trs: { ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2)) } 0.00/0.32 0.00/0.32 The induced complexity on above rules (modulo remaining rules) is 0.00/0.32 YES(?,O(n^1)) . These rules are moved into the corresponding weak 0.00/0.32 component(s). 0.00/0.32 0.00/0.32 Sub-proof: 0.00/0.32 ---------- 0.00/0.32 The following argument positions are usable: 0.00/0.32 Uargs(cons) = {2} 0.00/0.32 0.00/0.32 TcT has computed the following constructor-based matrix 0.00/0.32 interpretation satisfying not(EDA). 0.00/0.32 0.00/0.32 [is_empty](x1) = [5] 0.00/0.32 0.00/0.32 [nil] = [0] 0.00/0.32 0.00/0.32 [true] = [3] 0.00/0.32 0.00/0.32 [cons](x1, x2) = [1] x1 + [1] x2 + [4] 0.00/0.32 0.00/0.32 [false] = [5] 0.00/0.32 0.00/0.32 [hd](x1) = [3] x1 + [3] 0.00/0.32 0.00/0.32 [tl](x1) = [3] x1 + [3] 0.00/0.32 0.00/0.32 [append](x1, x2) = [2] x1 + [7] x2 + [3] 0.00/0.32 0.00/0.32 [ifappend](x1, x2, x3) = [7] x2 + [2] x3 + [0] 0.00/0.32 0.00/0.32 The order satisfies the following ordering constraints: 0.00/0.32 0.00/0.32 [is_empty(nil())] = [5] 0.00/0.32 > [3] 0.00/0.32 = [true()] 0.00/0.32 0.00/0.32 [is_empty(cons(x, l))] = [5] 0.00/0.32 >= [5] 0.00/0.32 = [false()] 0.00/0.32 0.00/0.32 [hd(cons(x, l))] = [3] x + [3] l + [15] 0.00/0.32 > [1] x + [0] 0.00/0.32 = [x] 0.00/0.32 0.00/0.32 [tl(cons(x, l))] = [3] x + [3] l + [15] 0.00/0.32 > [1] l + [0] 0.00/0.32 = [l] 0.00/0.32 0.00/0.32 [append(l1, l2)] = [2] l1 + [7] l2 + [3] 0.00/0.32 > [2] l1 + [7] l2 + [0] 0.00/0.32 = [ifappend(l1, l2, l1)] 0.00/0.32 0.00/0.32 [ifappend(l1, l2, nil())] = [7] l2 + [0] 0.00/0.32 >= [1] l2 + [0] 0.00/0.32 = [l2] 0.00/0.32 0.00/0.32 [ifappend(l1, l2, cons(x, l))] = [2] x + [2] l + [7] l2 + [8] 0.00/0.32 > [1] x + [2] l + [7] l2 + [7] 0.00/0.32 = [cons(x, append(l, l2))] 0.00/0.32 0.00/0.32 0.00/0.32 We return to the main proof. 0.00/0.32 0.00/0.32 We are left with following problem, upon which TcT provides the 0.00/0.32 certificate YES(O(1),O(1)). 0.00/0.32 0.00/0.32 Weak Trs: 0.00/0.32 { is_empty(nil()) -> true() 0.00/0.32 , is_empty(cons(x, l)) -> false() 0.00/0.32 , hd(cons(x, l)) -> x 0.00/0.32 , tl(cons(x, l)) -> l 0.00/0.32 , append(l1, l2) -> ifappend(l1, l2, l1) 0.00/0.32 , ifappend(l1, l2, nil()) -> l2 0.00/0.32 , ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2)) } 0.00/0.32 Obligation: 0.00/0.32 runtime complexity 0.00/0.32 Answer: 0.00/0.32 YES(O(1),O(1)) 0.00/0.32 0.00/0.32 Empty rules are trivially bounded 0.00/0.32 0.00/0.32 Hurray, we answered YES(O(1),O(n^1)) 0.00/0.32 EOF