We consider termination of the LCTRS with only rule scheme Calc:

  Signature: cond  :: Bool -> Int -> Int -> Int
             minus :: Int -> Int -> Int

  Rules: minus(x, y) -> cond(x >= y + 1, x, y)
         cond(false, x, y) -> 0
         cond(true, x, y) -> 1 + minus(x, y + 1)

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, f, c), where:

  P1. (1) minus#(x, y) => cond#(x >= y + 1, x, y)
      (2) cond#(true, x, y) => minus#(x, y + 1)

***** We apply the Theory Arguments Processor on D1 = (P1, R, f, c).

We use the following theory arguments function:

  cond#  : [1, 2, 3]
  minus# : [1, 2]

Processor output: { D2 = (P2, R, f, c) ; D3 = (P3, R, f, c) }, where:

  P2. (1) minus#(x, y) => cond#(x >= y + 1, x, y)
      (2) cond#(true, x, y) => minus#(x, y + 1) { x, y }

  P3. (1) cond#(true, x, y) => minus#(x, y + 1)

***** We apply the Theory Arguments Processor on D2 = (P2, R, f, c).

We use the following theory arguments function:

  cond#  : [1, 2, 3]
  minus# : [1, 2]

Processor output: { D4 = (P4, R, f, c) ; D5 = (P5, R, f, c) }, where:

  P4. (1) minus#(x, y) => cond#(x >= y + 1, x, y) { x, y }
      (2) cond#(true, x, y) => minus#(x, y + 1) { x, y }

  P5. (1) minus#(x, y) => cond#(x >= y + 1, x, y)

***** We apply the Graph Processor on D3 = (P3, 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: { }.

***** No progress could be made on DP problem D4 = (P4, R, f, c).