LTS Termination Proof

by AProVE

Input

• Initial Location: l5, l4, l6, l1, l0, l2
• Transitions: (pre-variables and post-variables)

  l0	1	l1:		x1 = _aHAT0 ∧ x2 = _bHAT0 ∧ x1 = _aHATpost ∧ x2 = _bHATpost ∧ _bHAT0 = _bHATpost ∧ _aHAT0 = _aHATpost ∧ 1 ≤ 0
  l1	2	l0:		x1 = _x ∧ x2 = _x1 ∧ x1 = _x2 ∧ x2 = _x3 ∧ _x1 = _x3 ∧ _x = _x2
  l0	3	l2:		x1 = _x4 ∧ x2 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ _x4 = _x6 ∧ _x7 = 0
  l2	4	l3:		x1 = _x8 ∧ x2 = _x9 ∧ x1 = _x10 ∧ x2 = _x11 ∧ _x9 = _x11 ∧ _x8 = _x10 ∧ 1 + _x9 ≤ 1
  l4	5	l0:		x1 = _x12 ∧ x2 = _x13 ∧ x1 = _x14 ∧ x2 = _x15 ∧ _x13 = _x15 ∧ _x12 = _x14 ∧ 1 + _x12 ≤ 1
  l4	6	l2:		x1 = _x16 ∧ x2 = _x17 ∧ x1 = _x18 ∧ x2 = _x19 ∧ _x16 = _x18 ∧ _x19 = 1
  l5	7	l4:		x1 = _x20 ∧ x2 = _x21 ∧ x1 = _x22 ∧ x2 = _x23 ∧ _x21 = _x23 ∧ _x22 = 1
  l6	8	l5:		x1 = _x24 ∧ x2 = _x25 ∧ x1 = _x26 ∧ x2 = _x27 ∧ _x25 = _x27 ∧ _x24 = _x26

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, 2 using the following ranking functions, which are bounded by −1.

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)