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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        implies#_A(x1,x2) = max{0, x2 - 2}
        not_A(x1) = max{4, x1}
        or_A(x1,x2) = max{3, x2 + 1}
  
    precedence:
    
      implies# = not = or
    
    partial status:
  
      pi(implies#) = []
      pi(not) = [1]
      pi(or) = [2]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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))

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}


-- 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))

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        implies#_A(x1,x2) = max{0, x1 - 3, x2 - 2}
        not_A(x1) = max{5, x1 + 2}
        or_A(x1,x2) = max{3, x1}
  
    precedence:
    
      implies# = not = or
    
    partial status:
  
      pi(implies#) = []
      pi(not) = [1]
      pi(or) = [1]

The next rules are strictly ordered:

  p1

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