We consider universal computability of the LCSTRS with only rule scheme Calc: Signature: l1 :: Int -> Int -> A (private) l2 :: Int -> Int -> A (private) l3 :: Int -> Int -> Int -> A (private) l4 :: Int -> Int -> Int -> A (private) l5 :: Int -> Int -> Int -> A (private) start :: Int -> A Rules: start(x) -> l1(x, 0) l1(x, y) -> l2(x, y) | x > y l2(x, y) -> l3(x, y, y) l3(x, y, z) -> l4(x, y, z) | z > 0 l4(x, y, z) -> l3(x, y, z - 1) l4(x, y, z) -> l5(x, y + 1, z) l5(x, y, z) -> l1(x, y) The system is accessible function passing by a sort ordering that equates all sorts. We start by computing the initial DP problem D1 = (P1, R UNION R_?, f, c), where: P1. (1) start#(x) => l1#(x, 0) (2) l1#(x, y) => l2#(x, y) | x > y (private) (3) l2#(x, y) => l3#(x, y, y) (private) (4) l3#(x, y, z) => l4#(x, y, z) | z > 0 (private) (5) l4#(x, y, z) => l3#(x, y, z - 1) (private) (6) l4#(x, y, z) => l5#(x, y + 1, z) (private) (7) l5#(x, y, z) => l1#(x, y) (private) ***** We apply the Theory Arguments Processor on D1 = (P1, R UNION R_?, f, c). We use the following theory arguments function, which fixes all public dependency pairs: l1# : [1, 2] l2# : [1, 2] l3# : [1, 2, 3] l4# : [1, 2, 3] l5# : [1, 2, 3] start# : [] Processor output: { D2 = (P2, R UNION R_?, f, c) }, where: P2. (1) start#(x) => l1#(x, 0) (2) l1#(x, y) => l2#(x, y) | x > y (private) (3) l2#(x, y) => l3#(x, y, y) { x, y } (private) (4) l3#(x, y, z) => l4#(x, y, z) | z > 0 { x, y, z } (private) (5) l4#(x, y, z) => l3#(x, y, z - 1) { x, y, z } (private) (6) l4#(x, y, z) => l5#(x, y + 1, z) { x, y, z } (private) (7) l5#(x, y, z) => l1#(x, y) { x, y, z } (private) ***** We apply the Graph Processor on D2 = (P2, R UNION R_?, f, c). We compute a graph approximation with the following edges: 1: 2 2: 3 3: 4 4: 5 6 5: 4 6: 7 7: 2 There is only one SCC, so all DPs not inside the SCC can be removed. Processor output: { D3 = (P3, R UNION R_?, f, c) }, where: P3. (1) l1#(x, y) => l2#(x, y) | x > y (2) l2#(x, y) => l3#(x, y, y) { x, y } (3) l3#(x, y, z) => l4#(x, y, z) | z > 0 { x, y, z } (4) l4#(x, y, z) => l3#(x, y, z - 1) { x, y, z } (5) l4#(x, y, z) => l5#(x, y + 1, z) { x, y, z } (6) l5#(x, y, z) => l1#(x, y) { x, y, z } ***** No progress could be made on DP problem D3 = (P3, R UNION R_?, f, c).