YES

We show the termination of the TRS R:

  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))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: implies#(not(x),or(y,z)) -> implies#(y,or(x,z))
p2: implies#(x,or(y,z)) -> implies#(x,z)

and R consists of:

r1: implies(not(x),y) -> or(x,y)
r2: implies(not(x),or(y,z)) -> implies(y,or(x,z))
r3: implies(x,or(y,z)) -> or(y,implies(x,z))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: implies#(not(x),or(y,z)) -> implies#(y,or(x,z))
p2: implies#(x,or(y,z)) -> implies#(x,z)

and R consists of:

r1: implies(not(x),y) -> or(x,y)
r2: implies(not(x),or(y,z)) -> implies(y,or(x,z))
r3: implies(x,or(y,z)) -> or(y,implies(x,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        implies#_A(x1,x2) = x1 + x2
        not_A(x1) = x1 + 1
        or_A(x1,x2) = x1 + x2 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        or > implies# > not
      
      argument filter:
    
        pi(implies#) = [2]
        pi(not) = []
        pi(or) = []
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.