*** 1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        active(and(X1,X2)) -> and(active(X1),X2)
        active(and(tt(),X)) -> mark(X)
        active(plus(N,0())) -> mark(N)
        active(plus(N,s(M))) -> mark(s(plus(N,M)))
        active(plus(X1,X2)) -> plus(X1,active(X2))
        active(plus(X1,X2)) -> plus(active(X1),X2)
        active(s(X)) -> s(active(X))
        and(mark(X1),X2) -> mark(and(X1,X2))
        and(ok(X1),ok(X2)) -> ok(and(X1,X2))
        plus(X1,mark(X2)) -> mark(plus(X1,X2))
        plus(mark(X1),X2) -> mark(plus(X1,X2))
        plus(ok(X1),ok(X2)) -> ok(plus(X1,X2))
        proper(0()) -> ok(0())
        proper(and(X1,X2)) -> and(proper(X1),proper(X2))
        proper(plus(X1,X2)) -> plus(proper(X1),proper(X2))
        proper(s(X)) -> s(proper(X))
        proper(tt()) -> ok(tt())
        s(mark(X)) -> mark(s(X))
        s(ok(X)) -> ok(s(X))
        top(mark(X)) -> top(proper(X))
        top(ok(X)) -> top(active(X))
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {active/1,and/2,plus/2,proper/1,s/1,top/1} / {0/0,mark/1,ok/1,tt/0}
      Obligation:
        Innermost
        basic terms: {active,and,plus,proper,s,top}/{0,mark,ok,tt}
    Applied Processor:
      Bounds {initialAutomaton = perSymbol, enrichment = match}
    Proof:
      The problem is match-bounded by 2.
      The enriched problem is compatible with follwoing automaton.
        0_0() -> 1
        0_1() -> 13
        active_0(1) -> 2
        active_0(4) -> 2
        active_0(5) -> 2
        active_0(10) -> 2
        active_1(1) -> 15
        active_1(4) -> 15
        active_1(5) -> 15
        active_1(10) -> 15
        active_2(13) -> 16
        and_0(1,1) -> 3
        and_0(1,4) -> 3
        and_0(1,5) -> 3
        and_0(1,10) -> 3
        and_0(4,1) -> 3
        and_0(4,4) -> 3
        and_0(4,5) -> 3
        and_0(4,10) -> 3
        and_0(5,1) -> 3
        and_0(5,4) -> 3
        and_0(5,5) -> 3
        and_0(5,10) -> 3
        and_0(10,1) -> 3
        and_0(10,4) -> 3
        and_0(10,5) -> 3
        and_0(10,10) -> 3
        and_1(1,1) -> 11
        and_1(1,4) -> 11
        and_1(1,5) -> 11
        and_1(1,10) -> 11
        and_1(4,1) -> 11
        and_1(4,4) -> 11
        and_1(4,5) -> 11
        and_1(4,10) -> 11
        and_1(5,1) -> 11
        and_1(5,4) -> 11
        and_1(5,5) -> 11
        and_1(5,10) -> 11
        and_1(10,1) -> 11
        and_1(10,4) -> 11
        and_1(10,5) -> 11
        and_1(10,10) -> 11
        mark_0(1) -> 4
        mark_0(4) -> 4
        mark_0(5) -> 4
        mark_0(10) -> 4
        mark_1(11) -> 3
        mark_1(11) -> 11
        mark_1(12) -> 6
        mark_1(12) -> 12
        mark_1(14) -> 8
        mark_1(14) -> 14
        ok_0(1) -> 5
        ok_0(4) -> 5
        ok_0(5) -> 5
        ok_0(10) -> 5
        ok_1(11) -> 3
        ok_1(11) -> 11
        ok_1(12) -> 6
        ok_1(12) -> 12
        ok_1(13) -> 7
        ok_1(13) -> 15
        ok_1(14) -> 8
        ok_1(14) -> 14
        plus_0(1,1) -> 6
        plus_0(1,4) -> 6
        plus_0(1,5) -> 6
        plus_0(1,10) -> 6
        plus_0(4,1) -> 6
        plus_0(4,4) -> 6
        plus_0(4,5) -> 6
        plus_0(4,10) -> 6
        plus_0(5,1) -> 6
        plus_0(5,4) -> 6
        plus_0(5,5) -> 6
        plus_0(5,10) -> 6
        plus_0(10,1) -> 6
        plus_0(10,4) -> 6
        plus_0(10,5) -> 6
        plus_0(10,10) -> 6
        plus_1(1,1) -> 12
        plus_1(1,4) -> 12
        plus_1(1,5) -> 12
        plus_1(1,10) -> 12
        plus_1(4,1) -> 12
        plus_1(4,4) -> 12
        plus_1(4,5) -> 12
        plus_1(4,10) -> 12
        plus_1(5,1) -> 12
        plus_1(5,4) -> 12
        plus_1(5,5) -> 12
        plus_1(5,10) -> 12
        plus_1(10,1) -> 12
        plus_1(10,4) -> 12
        plus_1(10,5) -> 12
        plus_1(10,10) -> 12
        proper_0(1) -> 7
        proper_0(4) -> 7
        proper_0(5) -> 7
        proper_0(10) -> 7
        proper_1(1) -> 15
        proper_1(4) -> 15
        proper_1(5) -> 15
        proper_1(10) -> 15
        s_0(1) -> 8
        s_0(4) -> 8
        s_0(5) -> 8
        s_0(10) -> 8
        s_1(1) -> 14
        s_1(4) -> 14
        s_1(5) -> 14
        s_1(10) -> 14
        top_0(1) -> 9
        top_0(4) -> 9
        top_0(5) -> 9
        top_0(10) -> 9
        top_1(15) -> 9
        top_2(16) -> 9
        tt_0() -> 10
        tt_1() -> 13
*** 1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        active(and(X1,X2)) -> and(active(X1),X2)
        active(and(tt(),X)) -> mark(X)
        active(plus(N,0())) -> mark(N)
        active(plus(N,s(M))) -> mark(s(plus(N,M)))
        active(plus(X1,X2)) -> plus(X1,active(X2))
        active(plus(X1,X2)) -> plus(active(X1),X2)
        active(s(X)) -> s(active(X))
        and(mark(X1),X2) -> mark(and(X1,X2))
        and(ok(X1),ok(X2)) -> ok(and(X1,X2))
        plus(X1,mark(X2)) -> mark(plus(X1,X2))
        plus(mark(X1),X2) -> mark(plus(X1,X2))
        plus(ok(X1),ok(X2)) -> ok(plus(X1,X2))
        proper(0()) -> ok(0())
        proper(and(X1,X2)) -> and(proper(X1),proper(X2))
        proper(plus(X1,X2)) -> plus(proper(X1),proper(X2))
        proper(s(X)) -> s(proper(X))
        proper(tt()) -> ok(tt())
        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:
        {active/1,and/2,plus/2,proper/1,s/1,top/1} / {0/0,mark/1,ok/1,tt/0}
      Obligation:
        Innermost
        basic terms: {active,and,plus,proper,s,top}/{0,mark,ok,tt}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).