*** 1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        f(x,y,z) -> g(<=(x,y),x,y,z)
        g(false(),x,y,z) -> f(f(p(x),y,z),f(p(y),z,x),f(p(z),x,y))
        g(true(),x,y,z) -> z
        p(0()) -> 0()
        p(s(x)) -> x
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {f/3,g/4,p/1} / {0/0,<=/2,false/0,s/1,true/0}
      Obligation:
        Innermost
        basic terms: {f,g,p}/{0,<=,false,s,true}
    Applied Processor:
      DependencyPairs {dpKind_ = DT}
    Proof:
      We add the following dependency tuples:
      
      Strict DPs
        f#(x,y,z) -> c_1(g#(<=(x,y),x,y,z))
        g#(false(),x,y,z) -> c_2(f#(f(p(x),y,z),f(p(y),z,x),f(p(z),x,y)),f#(p(x),y,z),p#(x),f#(p(y),z,x),p#(y),f#(p(z),x,y),p#(z))
        g#(true(),x,y,z) -> c_3()
        p#(0()) -> c_4()
        p#(s(x)) -> c_5()
      Weak DPs
        
      
      and mark the set of starting terms.
*** 1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        f#(x,y,z) -> c_1(g#(<=(x,y),x,y,z))
        g#(false(),x,y,z) -> c_2(f#(f(p(x),y,z),f(p(y),z,x),f(p(z),x,y)),f#(p(x),y,z),p#(x),f#(p(y),z,x),p#(y),f#(p(z),x,y),p#(z))
        g#(true(),x,y,z) -> c_3()
        p#(0()) -> c_4()
        p#(s(x)) -> c_5()
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        f(x,y,z) -> g(<=(x,y),x,y,z)
        g(false(),x,y,z) -> f(f(p(x),y,z),f(p(y),z,x),f(p(z),x,y))
        g(true(),x,y,z) -> z
        p(0()) -> 0()
        p(s(x)) -> x
      Signature:
        {f/3,g/4,p/1,f#/3,g#/4,p#/1} / {0/0,<=/2,false/0,s/1,true/0,c_1/1,c_2/7,c_3/0,c_4/0,c_5/0}
      Obligation:
        Innermost
        basic terms: {f#,g#,p#}/{0,<=,false,s,true}
    Applied Processor:
      UsableRules
    Proof:
      We replace rewrite rules by usable rules:
        f(x,y,z) -> g(<=(x,y),x,y,z)
        p(0()) -> 0()
        p(s(x)) -> x
        f#(x,y,z) -> c_1(g#(<=(x,y),x,y,z))
        g#(false(),x,y,z) -> c_2(f#(f(p(x),y,z),f(p(y),z,x),f(p(z),x,y)),f#(p(x),y,z),p#(x),f#(p(y),z,x),p#(y),f#(p(z),x,y),p#(z))
        g#(true(),x,y,z) -> c_3()
        p#(0()) -> c_4()
        p#(s(x)) -> c_5()
*** 1.1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        f#(x,y,z) -> c_1(g#(<=(x,y),x,y,z))
        g#(false(),x,y,z) -> c_2(f#(f(p(x),y,z),f(p(y),z,x),f(p(z),x,y)),f#(p(x),y,z),p#(x),f#(p(y),z,x),p#(y),f#(p(z),x,y),p#(z))
        g#(true(),x,y,z) -> c_3()
        p#(0()) -> c_4()
        p#(s(x)) -> c_5()
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        f(x,y,z) -> g(<=(x,y),x,y,z)
        p(0()) -> 0()
        p(s(x)) -> x
      Signature:
        {f/3,g/4,p/1,f#/3,g#/4,p#/1} / {0/0,<=/2,false/0,s/1,true/0,c_1/1,c_2/7,c_3/0,c_4/0,c_5/0}
      Obligation:
        Innermost
        basic terms: {f#,g#,p#}/{0,<=,false,s,true}
    Applied Processor:
      Trivial
    Proof:
      Consider the dependency graph
        1:S:f#(x,y,z) -> c_1(g#(<=(x,y),x,y,z))
           
        
        2:S:g#(false(),x,y,z) -> c_2(f#(f(p(x),y,z),f(p(y),z,x),f(p(z),x,y)),f#(p(x),y,z),p#(x),f#(p(y),z,x),p#(y),f#(p(z),x,y),p#(z))
           -->_7 p#(s(x)) -> c_5():5
           -->_5 p#(s(x)) -> c_5():5
           -->_3 p#(s(x)) -> c_5():5
           -->_7 p#(0()) -> c_4():4
           -->_5 p#(0()) -> c_4():4
           -->_3 p#(0()) -> c_4():4
           -->_6 f#(x,y,z) -> c_1(g#(<=(x,y),x,y,z)):1
           -->_4 f#(x,y,z) -> c_1(g#(<=(x,y),x,y,z)):1
           -->_2 f#(x,y,z) -> c_1(g#(<=(x,y),x,y,z)):1
           -->_1 f#(x,y,z) -> c_1(g#(<=(x,y),x,y,z)):1
        
        3:S:g#(true(),x,y,z) -> c_3()
           
        
        4:S:p#(0()) -> c_4()
           
        
        5:S:p#(s(x)) -> c_5()
           
        
      The dependency graph contains no loops, we remove all dependency pairs.
*** 1.1.1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        f(x,y,z) -> g(<=(x,y),x,y,z)
        p(0()) -> 0()
        p(s(x)) -> x
      Signature:
        {f/3,g/4,p/1,f#/3,g#/4,p#/1} / {0/0,<=/2,false/0,s/1,true/0,c_1/1,c_2/7,c_3/0,c_4/0,c_5/0}
      Obligation:
        Innermost
        basic terms: {f#,g#,p#}/{0,<=,false,s,true}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).