# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l5, l4, l7, l6, l11, l8, l1, l3, l0, l12, l2, l9
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = ___const_5HAT0 ∧ x2 = _i4HAT0 ∧ x3 = _j5HAT0 ∧ x4 = _k6HAT0 ∧ x1 = ___const_5HATpost ∧ x2 = _i4HATpost ∧ x3 = _j5HATpost ∧ x4 = _k6HATpost ∧ _k6HAT0 = _k6HATpost ∧ _j5HAT0 = _j5HATpost ∧ _i4HAT0 = _i4HATpost ∧ ___const_5HAT0 = ___const_5HATpost l2 2 l3: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x3 = _x7 ∧ _x2 = _x6 ∧ _x1 = _x5 ∧ _x = _x4 l4 3 l5: x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x11 = _x15 ∧ _x10 = _x14 ∧ _x9 = _x13 ∧ _x8 = _x12 l6 4 l7: x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ _x19 = _x23 ∧ _x18 = _x22 ∧ _x17 = _x21 ∧ _x16 = _x20 l8 5 l6: x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ _x27 = _x31 ∧ _x25 = _x29 ∧ _x24 = _x28 ∧ _x30 = 1 + _x26 ∧ 1 + _x24 ≤ _x27 l8 6 l9: x1 = _x32 ∧ x2 = _x33 ∧ x3 = _x34 ∧ x4 = _x35 ∧ x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ _x34 = _x38 ∧ _x33 = _x37 ∧ _x32 = _x36 ∧ _x39 = 1 + _x35 ∧ _x35 ≤ _x32 l7 7 l4: x1 = _x40 ∧ x2 = _x41 ∧ x3 = _x42 ∧ x4 = _x43 ∧ x1 = _x44 ∧ x2 = _x45 ∧ x3 = _x46 ∧ x4 = _x47 ∧ _x43 = _x47 ∧ _x42 = _x46 ∧ _x40 = _x44 ∧ _x45 = 1 + _x41 ∧ 1 + _x40 ≤ _x42 l7 8 l9: x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x4 = _x51 ∧ x1 = _x52 ∧ x2 = _x53 ∧ x3 = _x54 ∧ x4 = _x55 ∧ _x50 = _x54 ∧ _x49 = _x53 ∧ _x48 = _x52 ∧ _x55 = 1 ∧ _x50 ≤ _x48 l5 9 l10: x1 = _x56 ∧ x2 = _x57 ∧ x3 = _x58 ∧ x4 = _x59 ∧ x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ _x59 = _x63 ∧ _x58 = _x62 ∧ _x57 = _x61 ∧ _x56 = _x60 ∧ 1 + _x56 ≤ _x57 l5 10 l6: x1 = _x64 ∧ x2 = _x65 ∧ x3 = _x66 ∧ x4 = _x67 ∧ x1 = _x68 ∧ x2 = _x69 ∧ x3 = _x70 ∧ x4 = _x71 ∧ _x67 = _x71 ∧ _x65 = _x69 ∧ _x64 = _x68 ∧ _x70 = 1 ∧ _x65 ≤ _x64 l9 11 l8: x1 = _x72 ∧ x2 = _x73 ∧ x3 = _x74 ∧ x4 = _x75 ∧ x1 = _x76 ∧ x2 = _x77 ∧ x3 = _x78 ∧ x4 = _x79 ∧ _x75 = _x79 ∧ _x74 = _x78 ∧ _x73 = _x77 ∧ _x72 = _x76 l3 12 l0: x1 = _x80 ∧ x2 = _x81 ∧ x3 = _x82 ∧ x4 = _x83 ∧ x1 = _x84 ∧ x2 = _x85 ∧ x3 = _x86 ∧ x4 = _x87 ∧ _x83 = _x87 ∧ _x82 = _x86 ∧ _x80 = _x84 ∧ _x85 = 1 + _x81 ∧ 1 + _x80 ≤ _x82 l3 13 l2: x1 = _x88 ∧ x2 = _x89 ∧ x3 = _x90 ∧ x4 = _x91 ∧ x1 = _x92 ∧ x2 = _x93 ∧ x3 = _x94 ∧ x4 = _x95 ∧ _x91 = _x95 ∧ _x89 = _x93 ∧ _x88 = _x92 ∧ _x94 = 1 + _x90 ∧ _x90 ≤ _x88 l1 14 l4: x1 = _x96 ∧ x2 = _x97 ∧ x3 = _x98 ∧ x4 = _x99 ∧ x1 = _x100 ∧ x2 = _x101 ∧ x3 = _x102 ∧ x4 = _x103 ∧ _x99 = _x103 ∧ _x98 = _x102 ∧ _x96 = _x100 ∧ _x101 = 1 ∧ 1 + _x96 ≤ _x97 l1 15 l2: x1 = _x104 ∧ x2 = _x105 ∧ x3 = _x106 ∧ x4 = _x107 ∧ x1 = _x108 ∧ x2 = _x109 ∧ x3 = _x110 ∧ x4 = _x111 ∧ _x107 = _x111 ∧ _x105 = _x109 ∧ _x104 = _x108 ∧ _x110 = 1 ∧ _x105 ≤ _x104 l11 16 l0: x1 = _x112 ∧ x2 = _x113 ∧ x3 = _x114 ∧ x4 = _x115 ∧ x1 = _x116 ∧ x2 = _x117 ∧ x3 = _x118 ∧ x4 = _x119 ∧ _x115 = _x119 ∧ _x114 = _x118 ∧ _x112 = _x116 ∧ _x117 = 1 l12 17 l11: x1 = _x120 ∧ x2 = _x121 ∧ x3 = _x122 ∧ x4 = _x123 ∧ x1 = _x124 ∧ x2 = _x125 ∧ x3 = _x126 ∧ x4 = _x127 ∧ _x123 = _x127 ∧ _x122 = _x126 ∧ _x121 = _x125 ∧ _x120 = _x124

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l5 l5 l5: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l7 l7 l7: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l11 l11 l11: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l8 l8 l8: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l1 l1 l1: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l12 l12 l12: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l9 l9 l9: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
and for every transition t, a duplicate t is considered.

### 2 SCC Decomposition

We consider subproblems for each of the 2 SCC(s) of the program graph.

### 2.1 SCC Subproblem 1/2

Here we consider the SCC { l1, l3, l0, l2 }.

### 2.1.1 Transition Removal

We remove transition 15 using the following ranking functions, which are bounded by 0.

 l0: 4⋅x1 − 4⋅x2 + 1 l1: 4⋅x1 − 4⋅x2 l3: 4⋅x1 − 4⋅x2 − 1 l2: 4⋅x1 − 4⋅x2 − 1

### 2.1.2 Transition Removal

We remove transitions 1, 12 using the following ranking functions, which are bounded by 0.

 l0: 0 l1: −1 l3: 1 l2: 1

### 2.1.3 Transition Removal

We remove transition 13 using the following ranking functions, which are bounded by 0.

 l2: x1 − x3 l3: x1 − x3

### 2.1.4 Transition Removal

We remove transition 2 using the following ranking functions, which are bounded by 0.

 l2: 0 l3: −1

### 2.1.5 Trivial Cooperation Program

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

### 2.2 SCC Subproblem 2/2

Here we consider the SCC { l5, l4, l7, l6, l8, l9 }.

### 2.2.1 Transition Removal

We remove transition 10 using the following ranking functions, which are bounded by 0.

 l4: x1 − x2 l5: x1 − x2 l7: −1 + x1 − x2 l6: −1 + x1 − x2 l8: −1 + x1 − x2 l9: −1 + x1 − x2

### 2.2.2 Transition Removal

We remove transition 7 using the following ranking functions, which are bounded by 0.

 l4: −1 l5: −1 l7: 2 l6: 2 l8: 2 l9: 2

### 2.2.3 Transition Removal

We remove transition 3 using the following ranking functions, which are bounded by 0.

 l4: 0 l5: −1 l6: 4⋅x1 + 2⋅x2 − 3⋅x3 + 3 l7: 4⋅x1 + 2⋅x2 − 3⋅x3 + 2 l8: 4⋅x1 + 2⋅x2 − 3⋅x3 + 1 l9: 4⋅x1 + 2⋅x2 − 3⋅x3 + 1

### 2.2.4 Transition Removal

We remove transition 8 using the following ranking functions, which are bounded by 0.

 l6: 4⋅x1 − 4⋅x3 + 1 l7: 4⋅x1 − 4⋅x3 l8: 4⋅x1 − 4⋅x3 − 1 l9: 4⋅x1 − 4⋅x3 − 1

### 2.2.5 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by 0.

 l6: x1 − x4 l7: x1 − x4 l8: x1 − x4 l9: x1 − x4

### 2.2.6 Transition Removal

We remove transitions 4, 5, 11 using the following ranking functions, which are bounded by 0.

 l6: 0 l7: −1 l8: 1 l9: 2

### 2.2.7 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 % (16 real / 0 unknown / 0 assumptions / 16 total proof steps)