We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { half(0()) -> 0() , half(s(0())) -> 0() , half(s(s(x))) -> s(half(x)) , lastbit(0()) -> 0() , lastbit(s(0())) -> s(0()) , lastbit(s(s(x))) -> lastbit(x) , conv(0()) -> cons(nil(), 0()) , conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(conv) = {1}, Uargs(cons) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [half](x1) = [0] [0] = [0] [s](x1) = [1] x1 + [0] [lastbit](x1) = [4] [conv](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [nil] = [0] The order satisfies the following ordering constraints: [half(0())] = [0] >= [0] = [0()] [half(s(0()))] = [0] >= [0] = [0()] [half(s(s(x)))] = [0] >= [0] = [s(half(x))] [lastbit(0())] = [4] > [0] = [0()] [lastbit(s(0()))] = [4] > [0] = [s(0())] [lastbit(s(s(x)))] = [4] >= [4] = [lastbit(x)] [conv(0())] = [0] >= [0] = [cons(nil(), 0())] [conv(s(x))] = [1] x + [0] ? [4] = [cons(conv(half(s(x))), lastbit(s(x)))] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { half(0()) -> 0() , half(s(0())) -> 0() , half(s(s(x))) -> s(half(x)) , lastbit(s(s(x))) -> lastbit(x) , conv(0()) -> cons(nil(), 0()) , conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) } Weak Trs: { lastbit(0()) -> 0() , lastbit(s(0())) -> s(0()) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(conv) = {1}, Uargs(cons) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [half](x1) = [0] [0] = [1] [s](x1) = [1] x1 + [1] [lastbit](x1) = [1] x1 + [7] [conv](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [nil] = [7] The order satisfies the following ordering constraints: [half(0())] = [0] ? [1] = [0()] [half(s(0()))] = [0] ? [1] = [0()] [half(s(s(x)))] = [0] ? [1] = [s(half(x))] [lastbit(0())] = [8] > [1] = [0()] [lastbit(s(0()))] = [9] > [2] = [s(0())] [lastbit(s(s(x)))] = [1] x + [9] > [1] x + [7] = [lastbit(x)] [conv(0())] = [1] ? [8] = [cons(nil(), 0())] [conv(s(x))] = [1] x + [1] ? [1] x + [8] = [cons(conv(half(s(x))), lastbit(s(x)))] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { half(0()) -> 0() , half(s(0())) -> 0() , half(s(s(x))) -> s(half(x)) , conv(0()) -> cons(nil(), 0()) , conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) } Weak Trs: { lastbit(0()) -> 0() , lastbit(s(0())) -> s(0()) , lastbit(s(s(x))) -> lastbit(x) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(conv) = {1}, Uargs(cons) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [half](x1) = [1] [0] = [0] [s](x1) = [1] x1 + [0] [lastbit](x1) = [7] [conv](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [1] [nil] = [3] The order satisfies the following ordering constraints: [half(0())] = [1] > [0] = [0()] [half(s(0()))] = [1] > [0] = [0()] [half(s(s(x)))] = [1] >= [1] = [s(half(x))] [lastbit(0())] = [7] > [0] = [0()] [lastbit(s(0()))] = [7] > [0] = [s(0())] [lastbit(s(s(x)))] = [7] >= [7] = [lastbit(x)] [conv(0())] = [0] ? [4] = [cons(nil(), 0())] [conv(s(x))] = [1] x + [0] ? [9] = [cons(conv(half(s(x))), lastbit(s(x)))] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { half(s(s(x))) -> s(half(x)) , conv(0()) -> cons(nil(), 0()) , conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) } Weak Trs: { half(0()) -> 0() , half(s(0())) -> 0() , lastbit(0()) -> 0() , lastbit(s(0())) -> s(0()) , lastbit(s(s(x))) -> lastbit(x) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(conv) = {1}, Uargs(cons) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [half](x1) = [1] x1 + [1] [0] = [0] [s](x1) = [1] x1 + [0] [lastbit](x1) = [6] [conv](x1) = [1] x1 + [1] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [nil] = [0] The order satisfies the following ordering constraints: [half(0())] = [1] > [0] = [0()] [half(s(0()))] = [1] > [0] = [0()] [half(s(s(x)))] = [1] x + [1] >= [1] x + [1] = [s(half(x))] [lastbit(0())] = [6] > [0] = [0()] [lastbit(s(0()))] = [6] > [0] = [s(0())] [lastbit(s(s(x)))] = [6] >= [6] = [lastbit(x)] [conv(0())] = [1] > [0] = [cons(nil(), 0())] [conv(s(x))] = [1] x + [1] ? [1] x + [8] = [cons(conv(half(s(x))), lastbit(s(x)))] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { half(s(s(x))) -> s(half(x)) , conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) } Weak Trs: { half(0()) -> 0() , half(s(0())) -> 0() , lastbit(0()) -> 0() , lastbit(s(0())) -> s(0()) , lastbit(s(s(x))) -> lastbit(x) , conv(0()) -> cons(nil(), 0()) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(conv) = {1}, Uargs(cons) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [half](x1) = [1] x1 + [1] [0] = [0] [s](x1) = [1] x1 + [4] [lastbit](x1) = [6] [conv](x1) = [1] x1 + [1] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [nil] = [0] The order satisfies the following ordering constraints: [half(0())] = [1] > [0] = [0()] [half(s(0()))] = [5] > [0] = [0()] [half(s(s(x)))] = [1] x + [9] > [1] x + [5] = [s(half(x))] [lastbit(0())] = [6] > [0] = [0()] [lastbit(s(0()))] = [6] > [4] = [s(0())] [lastbit(s(s(x)))] = [6] >= [6] = [lastbit(x)] [conv(0())] = [1] > [0] = [cons(nil(), 0())] [conv(s(x))] = [1] x + [5] ? [1] x + [12] = [cons(conv(half(s(x))), lastbit(s(x)))] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) } Weak Trs: { half(0()) -> 0() , half(s(0())) -> 0() , half(s(s(x))) -> s(half(x)) , lastbit(0()) -> 0() , lastbit(s(0())) -> s(0()) , lastbit(s(s(x))) -> lastbit(x) , conv(0()) -> cons(nil(), 0()) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(s) = {1}, Uargs(conv) = {1}, Uargs(cons) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA) and not(IDA(1)). [half](x1) = [1 0] x1 + [0] [1 0] [1] [0] = [0] [1] [s](x1) = [1 0] x1 + [2] [1 0] [5] [lastbit](x1) = [0 0] x1 + [2] [1 0] [7] [conv](x1) = [1 2] x1 + [0] [1 0] [3] [cons](x1, x2) = [1 0] x1 + [1 0] x2 + [1] [0 0] [0 0] [0] [nil] = [1] [0] The order satisfies the following ordering constraints: [half(0())] = [0] [1] >= [0] [1] = [0()] [half(s(0()))] = [2] [3] > [0] [1] = [0()] [half(s(s(x)))] = [1 0] x + [4] [1 0] [5] > [1 0] x + [2] [1 0] [5] = [s(half(x))] [lastbit(0())] = [2] [7] > [0] [1] = [0()] [lastbit(s(0()))] = [2] [9] >= [2] [5] = [s(0())] [lastbit(s(s(x)))] = [0 0] x + [2] [1 0] [11] >= [0 0] x + [2] [1 0] [7] = [lastbit(x)] [conv(0())] = [2] [3] >= [2] [0] = [cons(nil(), 0())] [conv(s(x))] = [3 0] x + [12] [1 0] [5] > [3 0] x + [11] [0 0] [0] = [cons(conv(half(s(x))), lastbit(s(x)))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { half(0()) -> 0() , half(s(0())) -> 0() , half(s(s(x))) -> s(half(x)) , lastbit(0()) -> 0() , lastbit(s(0())) -> s(0()) , lastbit(s(s(x))) -> lastbit(x) , conv(0()) -> cons(nil(), 0()) , conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))