*** 1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        f(0(),y,0(),u) -> true()
        f(0(),y,s(z),u) -> false()
        f(s(x),0(),z,u) -> f(x,u,minus(z,s(x)),u)
        f(s(x),s(y),z,u) -> if(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))
        perfectp(0()) -> false()
        perfectp(s(x)) -> f(x,s(0()),s(x),s(x))
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {f/4,perfectp/1} / {0/0,false/0,if/3,le/2,minus/2,s/1,true/0}
      Obligation:
        Full
        basic terms: {f,perfectp}/{0,false,if,le,minus,s,true}
    Applied Processor:
      WeightGap {wgDimension = 1, wgDegree = 1, wgKind = Algebraic, wgUArgs = UArgs, wgOn = WgOnAny}
    Proof:
      The weightgap principle applies using the following nonconstant growth matrix-interpretation:
        We apply a matrix interpretation of kind constructor based matrix interpretation:
        The following argument positions are considered usable:
          uargs(if) = {3}
        
        Following symbols are considered usable:
          {}
        TcT has computed the following interpretation:
                 p(0) = [1]                  
                 p(f) = [6] x4 + [6]         
             p(false) = [0]                  
                p(if) = [1] x1 + [1] x3 + [4]
                p(le) = [0]                  
             p(minus) = [4]                  
          p(perfectp) = [8] x1 + [4]         
                 p(s) = [0]                  
              p(true) = [1]                  
        
        Following rules are strictly oriented:
         f(0(),y,0(),u) = [6] u + [6]
                        > [1]        
                        = true()     
        
        f(0(),y,s(z),u) = [6] u + [6]
                        > [0]        
                        = false()    
        
          perfectp(0()) = [12]       
                        > [0]        
                        = false()    
        
        
        Following rules are (at-least) weakly oriented:
         f(s(x),0(),z,u) =  [6] u + [6]              
                         >= [6] u + [6]              
                         =  f(x,u,minus(z,s(x)),u)   
        
        f(s(x),s(y),z,u) =  [6] u + [6]              
                         >= [6] u + [10]             
                         =  if(le(x,y)               
                              ,f(s(x),minus(y,x),z,u)
                              ,f(x,u,z,u))           
        
          perfectp(s(x)) =  [4]                      
                         >= [6]                      
                         =  f(x,s(0()),s(x),s(x))    
        
      Further, it can be verified that all rules not oriented are covered by the weightgap condition.
*** 1.1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        f(s(x),0(),z,u) -> f(x,u,minus(z,s(x)),u)
        f(s(x),s(y),z,u) -> if(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))
        perfectp(s(x)) -> f(x,s(0()),s(x),s(x))
      Weak DP Rules:
        
      Weak TRS Rules:
        f(0(),y,0(),u) -> true()
        f(0(),y,s(z),u) -> false()
        perfectp(0()) -> false()
      Signature:
        {f/4,perfectp/1} / {0/0,false/0,if/3,le/2,minus/2,s/1,true/0}
      Obligation:
        Full
        basic terms: {f,perfectp}/{0,false,if,le,minus,s,true}
    Applied Processor:
      WeightGap {wgDimension = 1, wgDegree = 1, wgKind = Algebraic, wgUArgs = UArgs, wgOn = WgOnAny}
    Proof:
      The weightgap principle applies using the following nonconstant growth matrix-interpretation:
        We apply a matrix interpretation of kind constructor based matrix interpretation:
        The following argument positions are considered usable:
          uargs(if) = {3}
        
        Following symbols are considered usable:
          {}
        TcT has computed the following interpretation:
                 p(0) = [0]          
                 p(f) = [4] x3 + [8] 
             p(false) = [1]          
                p(if) = [1] x3 + [12]
                p(le) = [1] x2 + [0] 
             p(minus) = [1] x1 + [5] 
          p(perfectp) = [8] x1 + [8] 
                 p(s) = [1]          
              p(true) = [1]          
        
        Following rules are strictly oriented:
        perfectp(s(x)) = [16]                 
                       > [12]                 
                       = f(x,s(0()),s(x),s(x))
        
        
        Following rules are (at-least) weakly oriented:
          f(0(),y,0(),u) =  [8]                      
                         >= [1]                      
                         =  true()                   
        
         f(0(),y,s(z),u) =  [12]                     
                         >= [1]                      
                         =  false()                  
        
         f(s(x),0(),z,u) =  [4] z + [8]              
                         >= [4] z + [28]             
                         =  f(x,u,minus(z,s(x)),u)   
        
        f(s(x),s(y),z,u) =  [4] z + [8]              
                         >= [4] z + [20]             
                         =  if(le(x,y)               
                              ,f(s(x),minus(y,x),z,u)
                              ,f(x,u,z,u))           
        
           perfectp(0()) =  [8]                      
                         >= [1]                      
                         =  false()                  
        
      Further, it can be verified that all rules not oriented are covered by the weightgap condition.
*** 1.1.1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        f(s(x),0(),z,u) -> f(x,u,minus(z,s(x)),u)
        f(s(x),s(y),z,u) -> if(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))
      Weak DP Rules:
        
      Weak TRS Rules:
        f(0(),y,0(),u) -> true()
        f(0(),y,s(z),u) -> false()
        perfectp(0()) -> false()
        perfectp(s(x)) -> f(x,s(0()),s(x),s(x))
      Signature:
        {f/4,perfectp/1} / {0/0,false/0,if/3,le/2,minus/2,s/1,true/0}
      Obligation:
        Full
        basic terms: {f,perfectp}/{0,false,if,le,minus,s,true}
    Applied Processor:
      NaturalMI {miDimension = 1, miDegree = 1, miKind = Algebraic, uargs = UArgs, urules = URules, selector = Just any strict-rules, greedy = NoGreedy}
    Proof:
      We apply a matrix interpretation of kind constructor based matrix interpretation:
      The following argument positions are considered usable:
        uargs(if) = {3}
      
      Following symbols are considered usable:
        {}
      TcT has computed the following interpretation:
               p(0) = [0]                  
               p(f) = [8] x1 + [1] x3 + [8]
           p(false) = [2]                  
              p(if) = [1] x3 + [8]         
              p(le) = [1] x2 + [1]         
           p(minus) = [12]                 
        p(perfectp) = [12] x1 + [2]        
               p(s) = [1] x1 + [2]         
            p(true) = [8]                  
      
      Following rules are strictly oriented:
       f(s(x),0(),z,u) = [8] x + [1] z + [24]     
                       > [8] x + [20]             
                       = f(x,u,minus(z,s(x)),u)   
      
      f(s(x),s(y),z,u) = [8] x + [1] z + [24]     
                       > [8] x + [1] z + [16]     
                       = if(le(x,y)               
                           ,f(s(x),minus(y,x),z,u)
                           ,f(x,u,z,u))           
      
      
      Following rules are (at-least) weakly oriented:
       f(0(),y,0(),u) =  [8]                  
                      >= [8]                  
                      =  true()               
      
      f(0(),y,s(z),u) =  [1] z + [10]         
                      >= [2]                  
                      =  false()              
      
        perfectp(0()) =  [2]                  
                      >= [2]                  
                      =  false()              
      
       perfectp(s(x)) =  [12] x + [26]        
                      >= [9] x + [10]         
                      =  f(x,s(0()),s(x),s(x))
      
*** 1.1.1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        f(0(),y,0(),u) -> true()
        f(0(),y,s(z),u) -> false()
        f(s(x),0(),z,u) -> f(x,u,minus(z,s(x)),u)
        f(s(x),s(y),z,u) -> if(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))
        perfectp(0()) -> false()
        perfectp(s(x)) -> f(x,s(0()),s(x),s(x))
      Signature:
        {f/4,perfectp/1} / {0/0,false/0,if/3,le/2,minus/2,s/1,true/0}
      Obligation:
        Full
        basic terms: {f,perfectp}/{0,false,if,le,minus,s,true}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).