YES Problem: g(A()) -> A() g(B()) -> A() g(B()) -> B() g(C()) -> A() g(C()) -> B() g(C()) -> C() foldB(t,0()) -> t foldB(t,s(n)) -> f(foldB(t,n),B()) foldC(t,0()) -> t foldC(t,s(n)) -> f(foldC(t,n),C()) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,s(c)) f'(triple(a,b,c),B()) -> f(triple(a,b,c),A()) f'(triple(a,b,c),A()) -> f''(foldB(triple(s(a),0(),c),b)) f''(triple(a,b,c)) -> foldC(triple(a,b,0()),c) Proof: Matrix Interpretation Processor: dim=1 interpretation: [f''](x0) = x0, [triple](x0, x1, x2) = x0 + 4x1 + 2x2, [f'](x0, x1) = x0 + x1 + 1, [foldC](x0, x1) = x0 + 2x1, [f](x0, x1) = x0 + x1 + 1, [s](x0) = x0 + 2, [foldB](x0, x1) = x0 + 4x1 + 1, [0] = 0, [C] = 3, [B] = 2, [g](x0) = x0, [A] = 2 orientation: g(A()) = 2 >= 2 = A() g(B()) = 2 >= 2 = A() g(B()) = 2 >= 2 = B() g(C()) = 3 >= 2 = A() g(C()) = 3 >= 2 = B() g(C()) = 3 >= 3 = C() foldB(t,0()) = t + 1 >= t = t foldB(t,s(n)) = 4n + t + 9 >= 4n + t + 4 = f(foldB(t,n),B()) foldC(t,0()) = t >= t = t foldC(t,s(n)) = 2n + t + 4 >= 2n + t + 4 = f(foldC(t,n),C()) f(t,x) = t + x + 1 >= t + x + 1 = f'(t,g(x)) f'(triple(a,b,c),C()) = a + 4b + 2c + 4 >= a + 4b + 2c + 4 = triple(a,b,s(c)) f'(triple(a,b,c),B()) = a + 4b + 2c + 3 >= a + 4b + 2c + 3 = f(triple(a,b,c),A()) f'(triple(a,b,c),A()) = a + 4b + 2c + 3 >= a + 4b + 2c + 3 = f''(foldB(triple(s(a),0(),c),b)) f''(triple(a,b,c)) = a + 4b + 2c >= a + 4b + 2c = foldC(triple(a,b,0()),c) problem: g(A()) -> A() g(B()) -> A() g(B()) -> B() g(C()) -> C() foldC(t,0()) -> t foldC(t,s(n)) -> f(foldC(t,n),C()) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,s(c)) f'(triple(a,b,c),B()) -> f(triple(a,b,c),A()) f'(triple(a,b,c),A()) -> f''(foldB(triple(s(a),0(),c),b)) f''(triple(a,b,c)) -> foldC(triple(a,b,0()),c) Matrix Interpretation Processor: dim=1 interpretation: [f''](x0) = x0 + 3, [triple](x0, x1, x2) = x0 + x1 + 2x2 + 6, [f'](x0, x1) = x0 + x1 + 4, [foldC](x0, x1) = x0 + 2x1 + 1, [f](x0, x1) = x0 + x1 + 4, [s](x0) = x0 + 4, [foldB](x0, x1) = x0 + x1, [0] = 1, [C] = 4, [B] = 4, [g](x0) = x0, [A] = 4 orientation: g(A()) = 4 >= 4 = A() g(B()) = 4 >= 4 = A() g(B()) = 4 >= 4 = B() g(C()) = 4 >= 4 = C() foldC(t,0()) = t + 3 >= t = t foldC(t,s(n)) = 2n + t + 9 >= 2n + t + 9 = f(foldC(t,n),C()) f(t,x) = t + x + 4 >= t + x + 4 = f'(t,g(x)) f'(triple(a,b,c),C()) = a + b + 2c + 14 >= a + b + 2c + 14 = triple(a,b,s(c)) f'(triple(a,b,c),B()) = a + b + 2c + 14 >= a + b + 2c + 14 = f(triple(a,b,c),A()) f'(triple(a,b,c),A()) = a + b + 2c + 14 >= a + b + 2c + 14 = f''(foldB(triple(s(a),0(),c),b)) f''(triple(a,b,c)) = a + b + 2c + 9 >= a + b + 2c + 9 = foldC(triple(a,b,0()),c) problem: g(A()) -> A() g(B()) -> A() g(B()) -> B() g(C()) -> C() foldC(t,s(n)) -> f(foldC(t,n),C()) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,s(c)) f'(triple(a,b,c),B()) -> f(triple(a,b,c),A()) f'(triple(a,b,c),A()) -> f''(foldB(triple(s(a),0(),c),b)) f''(triple(a,b,c)) -> foldC(triple(a,b,0()),c) Matrix Interpretation Processor: dim=1 interpretation: [f''](x0) = x0 + 5, [triple](x0, x1, x2) = x0 + x1 + x2 + 2, [f'](x0, x1) = x0 + x1 + 5, [foldC](x0, x1) = x0 + x1 + 1, [f](x0, x1) = x0 + x1 + 5, [s](x0) = x0 + 5, [foldB](x0, x1) = x0 + x1, [0] = 1, [C] = 0, [B] = 6, [g](x0) = x0, [A] = 6 orientation: g(A()) = 6 >= 6 = A() g(B()) = 6 >= 6 = A() g(B()) = 6 >= 6 = B() g(C()) = 0 >= 0 = C() foldC(t,s(n)) = n + t + 6 >= n + t + 6 = f(foldC(t,n),C()) f(t,x) = t + x + 5 >= t + x + 5 = f'(t,g(x)) f'(triple(a,b,c),C()) = a + b + c + 7 >= a + b + c + 7 = triple(a,b,s(c)) f'(triple(a,b,c),B()) = a + b + c + 13 >= a + b + c + 13 = f(triple(a,b,c),A()) f'(triple(a,b,c),A()) = a + b + c + 13 >= a + b + c + 13 = f''(foldB(triple(s(a),0(),c),b)) f''(triple(a,b,c)) = a + b + c + 7 >= a + b + c + 4 = foldC(triple(a,b,0()),c) problem: g(A()) -> A() g(B()) -> A() g(B()) -> B() g(C()) -> C() foldC(t,s(n)) -> f(foldC(t,n),C()) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,s(c)) f'(triple(a,b,c),B()) -> f(triple(a,b,c),A()) f'(triple(a,b,c),A()) -> f''(foldB(triple(s(a),0(),c),b)) Matrix Interpretation Processor: dim=1 interpretation: [f''](x0) = x0 + 2, [triple](x0, x1, x2) = x0 + 2x1 + 4x2 + 4, [f'](x0, x1) = x0 + x1 + 6, [foldC](x0, x1) = 4x0 + 7x1 + 2, [f](x0, x1) = x0 + 4x1 + 6, [s](x0) = x0 + 2, [foldB](x0, x1) = x0 + 2x1 + 2, [0] = 0, [C] = 2, [B] = 4, [g](x0) = 2x0, [A] = 0 orientation: g(A()) = 0 >= 0 = A() g(B()) = 8 >= 0 = A() g(B()) = 8 >= 4 = B() g(C()) = 4 >= 2 = C() foldC(t,s(n)) = 7n + 4t + 16 >= 7n + 4t + 16 = f(foldC(t,n),C()) f(t,x) = t + 4x + 6 >= t + 2x + 6 = f'(t,g(x)) f'(triple(a,b,c),C()) = a + 2b + 4c + 12 >= a + 2b + 4c + 12 = triple(a,b,s(c)) f'(triple(a,b,c),B()) = a + 2b + 4c + 14 >= a + 2b + 4c + 10 = f(triple(a,b,c),A()) f'(triple(a,b,c),A()) = a + 2b + 4c + 10 >= a + 2b + 4c + 10 = f''(foldB(triple(s(a),0(),c),b)) problem: g(A()) -> A() foldC(t,s(n)) -> f(foldC(t,n),C()) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,s(c)) f'(triple(a,b,c),A()) -> f''(foldB(triple(s(a),0(),c),b)) DP Processor: DPs: foldC#(t,s(n)) -> foldC#(t,n) foldC#(t,s(n)) -> f#(foldC(t,n),C()) f#(t,x) -> g#(x) f#(t,x) -> f'#(t,g(x)) TRS: g(A()) -> A() foldC(t,s(n)) -> f(foldC(t,n),C()) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,s(c)) f'(triple(a,b,c),A()) -> f''(foldB(triple(s(a),0(),c),b)) TDG Processor: DPs: foldC#(t,s(n)) -> foldC#(t,n) foldC#(t,s(n)) -> f#(foldC(t,n),C()) f#(t,x) -> g#(x) f#(t,x) -> f'#(t,g(x)) TRS: g(A()) -> A() foldC(t,s(n)) -> f(foldC(t,n),C()) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,s(c)) f'(triple(a,b,c),A()) -> f''(foldB(triple(s(a),0(),c),b)) graph: foldC#(t,s(n)) -> f#(foldC(t,n),C()) -> f#(t,x) -> f'#(t,g(x)) foldC#(t,s(n)) -> f#(foldC(t,n),C()) -> f#(t,x) -> g#(x) foldC#(t,s(n)) -> foldC#(t,n) -> foldC#(t,s(n)) -> f#(foldC(t,n),C()) foldC#(t,s(n)) -> foldC#(t,n) -> foldC#(t,s(n)) -> foldC#(t,n) SCC Processor: #sccs: 1 #rules: 1 #arcs: 4/16 DPs: foldC#(t,s(n)) -> foldC#(t,n) TRS: g(A()) -> A() foldC(t,s(n)) -> f(foldC(t,n),C()) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,s(c)) f'(triple(a,b,c),A()) -> f''(foldB(triple(s(a),0(),c),b)) Subterm Criterion Processor: simple projection: pi(foldC#) = 1 problem: DPs: TRS: g(A()) -> A() foldC(t,s(n)) -> f(foldC(t,n),C()) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,s(c)) f'(triple(a,b,c),A()) -> f''(foldB(triple(s(a),0(),c),b)) Qed