LTS Termination Proof

by AProVE

Input

• Initial Location: l4, l1, l3, l0
• Transitions: (pre-variables and post-variables)

  l0	1	l1:		x1 = _Result_4HAT0 ∧ x2 = _b_7HAT0 ∧ x3 = _x_5HAT0 ∧ x4 = _y_6HAT0 ∧ x1 = _Result_4HATpost ∧ x2 = _b_7HATpost ∧ x3 = _x_5HATpost ∧ x4 = _y_6HATpost ∧ _x_5HAT0 = _x_5HATpost ∧ _Result_4HAT0 = _Result_4HATpost ∧ _y_6HATpost = −1 + _y_6HAT0 ∧ _b_7HATpost = 0 ∧ 1 − _b_7HAT0 ≤ 0 ∧ 0 ≤ −1 − _x_5HAT0 + _y_6HAT0
  l0	2	l2:		x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x3 = _x7 ∧ _x2 = _x6 ∧ _x1 = _x5 ∧ _x4 = _x4 ∧ − _x2 + _x3 ≤ 0
  l3	3	l1:		x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x11 = _x15 ∧ _x10 = _x14 ∧ _x8 = _x12 ∧ _x13 = 0
  l1	4	l0:		x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ _x19 = _x23 ∧ _x16 = _x20 ∧ _x22 = 1 + _x18 ∧ _x21 = 1 ∧ 0 ≤ − _x17 ∧ 0 ≤ −1 − _x18 + _x19
  l1	5	l2:		x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ _x27 = _x31 ∧ _x26 = _x30 ∧ _x25 = _x29 ∧ _x28 = _x28 ∧ − _x26 + _x27 ≤ 0
  l4	6	l3:		x1 = _x32 ∧ x2 = _x33 ∧ x3 = _x34 ∧ x4 = _x35 ∧ x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ _x35 = _x39 ∧ _x34 = _x38 ∧ _x33 = _x37 ∧ _x32 = _x36

Proof

1 Switch to Cooperation Termination Proof

2 SCC Decomposition

We consider subproblems for each of the 1 SCC(s) of the program graph.

2.1 SCC Subproblem 1/1

Here we consider the SCC { l1, l0 }.

2.1.1 Transition Removal

We remove transitions 1, 4 using the following ranking functions, which are bounded by 0.

2.1.2 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

Tool configuration

AProVE

• version: AProVE Commit ID: unknown
• strategy: Statistics for single proof: 100.00 % (4 real / 0 unknown / 0 assumptions / 4 total proof steps)