YES
Problem:
strict:
R(x,B2()) -> B2()
W(x,B2()) -> B2()
weak:
B1() -> R(T(),B1())
B1() -> W(T(),B1())
Proof:
Matrix Interpretation Processor: dim=5
interpretation:
[0]
[0]
[T] = [0]
[0]
[0],
[1 0 0 0 0] [1 0 0 0 1]
[0 0 0 0 0] [0 0 1 0 1]
[R](x0, x1) = [0 0 0 0 0]x0 + [1 1 0 0 0]x1
[0 0 0 0 0] [0 0 0 0 1]
[0 0 0 0 0] [0 0 0 1 1] ,
[0]
[0]
[B1] = [0]
[0]
[0],
[1]
[1]
[B2] = [1]
[1]
[1],
[1 0 0 0 0] [1 0 0 1 1]
[0 0 0 0 0] [1 0 1 0 0]
[W](x0, x1) = [0 0 0 0 0]x0 + [0 0 0 0 1]x1
[0 0 0 0 0] [1 0 1 0 0]
[0 0 0 0 0] [0 0 0 0 1]
orientation:
[1 0 0 0 0] [2] [1]
[0 0 0 0 0] [2] [1]
R(x,B2()) = [0 0 0 0 0]x + [2] >= [1] = B2()
[0 0 0 0 0] [1] [1]
[0 0 0 0 0] [2] [1]
[1 0 0 0 0] [3] [1]
[0 0 0 0 0] [2] [1]
W(x,B2()) = [0 0 0 0 0]x + [1] >= [1] = B2()
[0 0 0 0 0] [2] [1]
[0 0 0 0 0] [1] [1]
[0] [0]
[0] [0]
B1() = [0] >= [0] = R(T(),B1())
[0] [0]
[0] [0]
[0] [0]
[0] [0]
B1() = [0] >= [0] = W(T(),B1())
[0] [0]
[0] [0]
problem:
strict:
weak:
B1() -> R(T(),B1())
B1() -> W(T(),B1())
Qed