YES

Problem:
 implies(not(x),y) -> or(x,y)
 implies(not(x),or(y,z)) -> implies(y,or(x,z))
 implies(x,or(y,z)) -> or(y,implies(x,z))

Proof:
 DP Processor:
  DPs:
   implies#(not(x),or(y,z)) -> implies#(y,or(x,z))
   implies#(x,or(y,z)) -> implies#(x,z)
  TRS:
   implies(not(x),y) -> or(x,y)
   implies(not(x),or(y,z)) -> implies(y,or(x,z))
   implies(x,or(y,z)) -> or(y,implies(x,z))
  Usable Rule Processor:
   DPs:
    implies#(not(x),or(y,z)) -> implies#(y,or(x,z))
    implies#(x,or(y,z)) -> implies#(x,z)
   TRS:
    
   Matrix Interpretation Processor: dim=4
    
    usable rules:
     
    interpretation:
     [implies#](x0, x1) = [0 0 1 1]x0 + [1 0 1 1]x1,
     
                    [0 0 0 0]     [1 0 1 1]     [1]
                    [0 0 0 0]     [0 0 0 0]     [0]
     [or](x0, x1) = [0 0 0 1]x0 + [0 0 1 1]x1 + [1]
                    [0 0 1 1]     [1 1 0 0]     [0],
     
                 [0 0 0 0]     [0]
                 [0 0 0 0]     [0]
     [not](x0) = [0 0 1 1]x0 + [1]
                 [0 0 0 1]     [0]
    orientation:
     implies#(not(x),or(y,z)) = [0 0 1 2]x + [0 0 1 2]y + [2 1 2 2]z + [3] >= [0 0 1 2]x + [0 0 1 1]y + [2 1 2 2]z + [2] = implies#(y,or(x,z))
     
     implies#(x,or(y,z)) = [0 0 1 1]x + [0 0 1 2]y + [2 1 2 2]z + [2] >= [0 0 1 1]x + [1 0 1 1]z = implies#(x,z)
    problem:
     DPs:
      
     TRS:
      
    Qed