Problem:
 active(and(tt(),X)) -> mark(X)
 active(plus(N,0())) -> mark(N)
 active(plus(N,s(M))) -> mark(s(plus(N,M)))
 mark(and(X1,X2)) -> active(and(mark(X1),X2))
 mark(tt()) -> active(tt())
 mark(plus(X1,X2)) -> active(plus(mark(X1),mark(X2)))
 mark(0()) -> active(0())
 mark(s(X)) -> active(s(mark(X)))
 and(mark(X1),X2) -> and(X1,X2)
 and(X1,mark(X2)) -> and(X1,X2)
 and(active(X1),X2) -> and(X1,X2)
 and(X1,active(X2)) -> and(X1,X2)
 plus(mark(X1),X2) -> plus(X1,X2)
 plus(X1,mark(X2)) -> plus(X1,X2)
 plus(active(X1),X2) -> plus(X1,X2)
 plus(X1,active(X2)) -> plus(X1,X2)
 s(mark(X)) -> s(X)
 s(active(X)) -> s(X)

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {7,6,5,4,3}
   transitions:
    active1(9) -> 4*
    01() -> 9*
    tt1() -> 9*
    active0(2) -> 3*
    active0(1) -> 3*
    and0(1,2) -> 5*
    and0(2,1) -> 5*
    and0(1,1) -> 5*
    and0(2,2) -> 5*
    tt0() -> 1*
    mark0(2) -> 4*
    mark0(1) -> 4*
    plus0(1,2) -> 6*
    plus0(2,1) -> 6*
    plus0(1,1) -> 6*
    plus0(2,2) -> 6*
    00() -> 2*
    s0(2) -> 7*
    s0(1) -> 7*
  problem:
   
  Qed