LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l4, l3, l0, l2

Transitions: (pre-variables and post-variables)

l0	1	l1:		x1 = ____len9HAT0 ∧ x2 = ____val8HAT0 ∧ x3 = _tmp10HAT0 ∧ x4 = _tmpHAT0 ∧ x5 = _tmp___0HAT0 ∧ x6 = _tmp___1HAT0 ∧ x1 = ____len9HATpost ∧ x2 = ____val8HATpost ∧ x3 = _tmp10HATpost ∧ x4 = _tmpHATpost ∧ x5 = _tmp___0HATpost ∧ x6 = _tmp___1HATpost ∧ _tmp___1HAT0 = _tmp___1HATpost ∧ _tmp___0HAT0 = _tmp___0HATpost ∧ _tmp10HAT0 = _tmp10HATpost ∧ _tmpHAT0 = _tmpHATpost ∧ ____val8HAT0 = ____val8HATpost ∧ ____len9HAT0 = ____len9HATpost
l2	2	l0:		x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x6 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x4 = _x9 ∧ x5 = _x10 ∧ x6 = _x11 ∧ _x5 = _x11 ∧ _x2 = _x8 ∧ _x3 = _x9 ∧ _x1 = _x7 ∧ _x = _x6 ∧ _x10 = _x10
l3	3	l0:		x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ x5 = _x16 ∧ x6 = _x17 ∧ x1 = _x18 ∧ x2 = _x19 ∧ x3 = _x20 ∧ x4 = _x21 ∧ x5 = _x22 ∧ x6 = _x23 ∧ _x17 = _x23 ∧ _x16 = _x22 ∧ _x15 = _x21 ∧ _x20 = _x20 ∧ _x18 = _x18 ∧ _x19 = 0 ∧ −1 ≤ _x17 ∧ _x17 ≤ −1
l3	4	l2:		x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x5 = _x28 ∧ x6 = _x29 ∧ x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x4 = _x33 ∧ x5 = _x34 ∧ x6 = _x35 ∧ _x29 = _x35 ∧ _x28 = _x34 ∧ _x26 = _x32 ∧ _x27 = _x33 ∧ _x25 = _x31 ∧ _x24 = _x30 ∧ 0 ≤ _x29
l3	5	l2:		x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ x5 = _x40 ∧ x6 = _x41 ∧ x1 = _x42 ∧ x2 = _x43 ∧ x3 = _x44 ∧ x4 = _x45 ∧ x5 = _x46 ∧ x6 = _x47 ∧ _x41 = _x47 ∧ _x40 = _x46 ∧ _x38 = _x44 ∧ _x39 = _x45 ∧ _x37 = _x43 ∧ _x36 = _x42 ∧ 1 + _x41 ≤ −1
l4	6	l3:		x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x4 = _x51 ∧ x5 = _x52 ∧ x6 = _x53 ∧ x1 = _x54 ∧ x2 = _x55 ∧ x3 = _x56 ∧ x4 = _x57 ∧ x5 = _x58 ∧ x6 = _x59 ∧ _x52 = _x58 ∧ _x50 = _x56 ∧ _x49 = _x55 ∧ _x48 = _x54 ∧ _x59 = _x59 ∧ _x57 = _x57
l5	7	l4:		x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ x5 = _x64 ∧ x6 = _x65 ∧ x1 = _x66 ∧ x2 = _x67 ∧ x3 = _x68 ∧ x4 = _x69 ∧ x5 = _x70 ∧ x6 = _x71 ∧ _x65 = _x71 ∧ _x64 = _x70 ∧ _x62 = _x68 ∧ _x63 = _x69 ∧ _x61 = _x67 ∧ _x60 = _x66

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

l5	l5	l5:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6
l4	l4	l4:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6
l3	l3	l3:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6
l2	l2	l2:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6

and for every transition t, a duplicate t is considered.

2 SCC Decomposition

There exist no SCC in the program graph.

Tool configuration

version: AProVE Commit ID: unknown
strategy: Statistics for single proof: 100.00 % (2 real / 0 unknown / 0 assumptions / 2 total proof steps)