We consider universal computability of the LCTRS with only rule scheme Calc: Signature: eval_1 :: Int -> Int -> Int -> o eval_2 :: Int -> Int -> Int -> o Rules: eval_1(x, y, z) -> eval_2(x, y, z) | x > z eval_2(x, y, z) -> eval_2(x, y - 1, z) | x > z /\ y > z eval_2(x, y, z) -> eval_1(x - 1, y, z) | x > z /\ z >= 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) eval_1#(x, y, z) => eval_2#(x, y, z) | x > z (2) eval_2#(x, y, z) => eval_2#(x, y - 1, z) | x > z /\ y > z (3) eval_2#(x, y, z) => eval_1#(x - 1, y, z) | x > z /\ z >= y ***** We apply the Theory Arguments Processor on D1 = (P1, R UNION R_?, f, c). We use the following theory arguments function: eval_2# : [1, 2, 3] Processor output: { D2 = (P2, R UNION R_?, f, c) ; D3 = (P3, R UNION R_?, f, c) }, where: P2. (1) eval_1#(x, y, z) => eval_2#(x, y, z) | x > z { x, y, z } (2) eval_2#(x, y, z) => eval_2#(x, y - 1, z) | x > z /\ y > z (3) eval_2#(x, y, z) => eval_1#(x - 1, y, z) | x > z /\ z >= y P3. (1) eval_1#(x, y, z) => eval_2#(x, y, z) | x > z ***** We apply the Integer Function Processor on D2 = (P2, R UNION R_?, f, c). We use the following integer mapping: J(eval_1#) = arg_1 - arg_3 - 1 J(eval_2#) = arg_1 - arg_3 - 1 We thus have: (1) x > z |= x - z - 1 >= x - z - 1 (2) x > z /\ y > z |= x - z - 1 >= x - z - 1 (3) x > z /\ z >= y |= x - z - 1 > x - 1 - z - 1 (and x - z - 1 >= 0) We may remove the strictly oriented DPs, which yields: Processor output: { D4 = (P4, R UNION R_?, f, c) }, where: P4. (1) eval_1#(x, y, z) => eval_2#(x, y, z) | x > z { x, y, z } (2) eval_2#(x, y, z) => eval_2#(x, y - 1, z) | x > z /\ y > z ***** We apply the Graph Processor on D3 = (P3, R UNION R_?, f, c). We compute a graph approximation with the following edges: 1: As there are no SCCs, this DP problem is removed. Processor output: { }. ***** We apply the Graph Processor on D4 = (P4, R UNION R_?, f, c). We compute a graph approximation with the following edges: 1: 2 2: 2 There is only one SCC, so all DPs not inside the SCC can be removed. Processor output: { D5 = (P5, R UNION R_?, f, c) }, where: P5. (1) eval_2#(x, y, z) => eval_2#(x, y - 1, z) | x > z /\ y > z ***** We apply the Integer Function Processor on D5 = (P5, R UNION R_?, f, c). We use the following integer mapping: J(eval_2#) = arg_2 - arg_3 We thus have: (1) x > z /\ y > z |= y - z > y - 1 - z (and y - z >= 0) All DPs are strictly oriented, and may be removed. Hence, this DP problem is finite. Processor output: { }.