* Step 1: Bounds WORST_CASE(?,O(n^1)) + Considered Problem: - Strict TRS: diff(X1,mark(X2)) -> mark(diff(X1,X2)) diff(mark(X1),X2) -> mark(diff(X1,X2)) diff(ok(X1),ok(X2)) -> ok(diff(X1,X2)) if(mark(X1),X2,X3) -> mark(if(X1,X2,X3)) if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3)) leq(X1,mark(X2)) -> mark(leq(X1,X2)) leq(mark(X1),X2) -> mark(leq(X1,X2)) leq(ok(X1),ok(X2)) -> ok(leq(X1,X2)) p(mark(X)) -> mark(p(X)) p(ok(X)) -> ok(p(X)) proper(0()) -> ok(0()) proper(false()) -> ok(false()) proper(true()) -> ok(true()) s(mark(X)) -> mark(s(X)) s(ok(X)) -> ok(s(X)) top(mark(X)) -> top(proper(X)) top(ok(X)) -> top(active(X)) - Signature: {diff/2,if/3,leq/2,p/1,proper/1,s/1,top/1} / {0/0,active/1,false/0,mark/1,ok/1,true/0} - Obligation: innermost runtime complexity wrt. defined symbols {diff,if,leq,p,proper,s,top} and constructors {0,active ,false,mark,ok,true} + Applied Processor: Bounds {initialAutomaton = minimal, enrichment = match} + Details: The problem is match-bounded by 2. The enriched problem is compatible with follwoing automaton. 0_0() -> 2 0_1() -> 3 active_0(2) -> 2 active_1(2) -> 4 active_2(3) -> 5 diff_0(2,2) -> 1 diff_1(2,2) -> 3 false_0() -> 2 false_1() -> 3 if_0(2,2,2) -> 1 if_1(2,2,2) -> 3 leq_0(2,2) -> 1 leq_1(2,2) -> 3 mark_0(2) -> 2 mark_1(3) -> 1 mark_1(3) -> 3 ok_0(2) -> 2 ok_1(3) -> 1 ok_1(3) -> 3 ok_1(3) -> 4 p_0(2) -> 1 p_1(2) -> 3 proper_0(2) -> 1 proper_1(2) -> 4 s_0(2) -> 1 s_1(2) -> 3 top_0(2) -> 1 top_1(4) -> 1 top_2(5) -> 1 true_0() -> 2 true_1() -> 3 * Step 2: EmptyProcessor WORST_CASE(?,O(1)) + Considered Problem: - Weak TRS: diff(X1,mark(X2)) -> mark(diff(X1,X2)) diff(mark(X1),X2) -> mark(diff(X1,X2)) diff(ok(X1),ok(X2)) -> ok(diff(X1,X2)) if(mark(X1),X2,X3) -> mark(if(X1,X2,X3)) if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3)) leq(X1,mark(X2)) -> mark(leq(X1,X2)) leq(mark(X1),X2) -> mark(leq(X1,X2)) leq(ok(X1),ok(X2)) -> ok(leq(X1,X2)) p(mark(X)) -> mark(p(X)) p(ok(X)) -> ok(p(X)) proper(0()) -> ok(0()) proper(false()) -> ok(false()) proper(true()) -> ok(true()) s(mark(X)) -> mark(s(X)) s(ok(X)) -> ok(s(X)) top(mark(X)) -> top(proper(X)) top(ok(X)) -> top(active(X)) - Signature: {diff/2,if/3,leq/2,p/1,proper/1,s/1,top/1} / {0/0,active/1,false/0,mark/1,ok/1,true/0} - Obligation: innermost runtime complexity wrt. defined symbols {diff,if,leq,p,proper,s,top} and constructors {0,active ,false,mark,ok,true} + Applied Processor: EmptyProcessor + Details: The problem is already closed. The intended complexity is O(1). WORST_CASE(?,O(n^1))