# LTS Termination Proof

by T2Cert

## Input

Integer Transition System
• Initial Location: 3
• Transitions: (pre-variables and post-variables)  0 0 1: 1 − x_0 ≤ 0 ∧ 1 − y_0 ≤ 0 ∧ − z_0 + z_0 ≤ 0 ∧ z_0 − z_0 ≤ 0 ∧ − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 1 1 0: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − y_0 + y_post ≤ 0 ∧ −1 + y_0 − y_post ≤ 0 ∧ x_post − z_0 ≤ 0 ∧ − x_post + z_0 ≤ 0 ∧ x_0 − x_post ≤ 0 ∧ − x_0 + x_post ≤ 0 ∧ y_0 − y_post ≤ 0 ∧ − y_0 + y_post ≤ 0 ∧ − z_0 + z_0 ≤ 0 ∧ z_0 − z_0 ≤ 0 1 2 0: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − x_0 + x_post ≤ 0 ∧ −1 + x_0 − x_post ≤ 0 ∧ x_0 − x_post ≤ 0 ∧ − x_0 + x_post ≤ 0 ∧ − z_0 + z_0 ≤ 0 ∧ z_0 − z_0 ≤ 0 ∧ − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 2 3 0: − z_0 + z_0 ≤ 0 ∧ z_0 − z_0 ≤ 0 ∧ − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 3 4 2: − z_0 + z_0 ≤ 0 ∧ z_0 − z_0 ≤ 0 ∧ − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0

## Proof

The following invariants are asserted.

 0: TRUE 1: 1 − x_0 ≤ 0 ∧ 1 − y_0 ≤ 0 2: TRUE 3: TRUE

The invariants are proved as follows.

### IMPACT Invariant Proof

• nodes (location) invariant:  0 (0) TRUE 1 (1) 1 − x_0 ≤ 0 ∧ 1 − y_0 ≤ 0 2 (2) TRUE 3 (3) TRUE
• initial node: 3
• cover edges:
• transition edges:  0 0 1 1 1 0 1 2 0 2 3 0 3 4 2

### 2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 0 5 0: − z_0 + z_0 ≤ 0 ∧ z_0 − z_0 ≤ 0 ∧ − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0
and for every transition t, a duplicate t is considered.

### 3 Transition Removal

We remove transitions 3, 4 using the following ranking functions, which are bounded by −11.

 3: 0 2: 0 0: 0 1: 0 3: −4 2: −5 0: −6 1: −6 0_var_snapshot: −6 0*: −6
Hints:
 6 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 0 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 1 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 2 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 3 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 4 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

0* 8 0: z_0 + z_0 ≤ 0z_0z_0 ≤ 0y_post + y_post ≤ 0y_posty_post ≤ 0y_0 + y_0 ≤ 0y_0y_0 ≤ 0x_post + x_post ≤ 0x_postx_post ≤ 0x_0 + x_0 ≤ 0x_0x_0 ≤ 0

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

0 6 0_var_snapshot: z_0 + z_0 ≤ 0z_0z_0 ≤ 0y_post + y_post ≤ 0y_posty_post ≤ 0y_0 + y_0 ≤ 0y_0y_0 ≤ 0x_post + x_post ≤ 0x_postx_post ≤ 0x_0 + x_0 ≤ 0x_0x_0 ≤ 0

### 6 SCC Decomposition

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

### 6.1 SCC Subproblem 1/1

Here we consider the SCC { 0, 1, 0_var_snapshot, 0* }.

### 6.1.1 Transition Removal

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

 0: y_0 1: y_0 0_var_snapshot: y_0 0*: y_0
Hints:
 6 lexWeak[ [0, 0, 0, 0, 1, 0, 0, 0, 0, 0] ] 8 lexWeak[ [0, 0, 0, 0, 1, 0, 0, 0, 0, 0] ] 0 lexWeak[ [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0] ] 1 lexStrict[ [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0] , [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 2 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0] ]

### 6.1.2 Transition Removal

We remove transitions 0, 2 using the following ranking functions, which are bounded by 3.

 0: 2 + 4⋅x_0 1: 4⋅x_0 0_var_snapshot: 1 + 4⋅x_0 0*: 3 + 4⋅x_0
Hints:
 6 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 4, 0] ] 8 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 4, 0] ] 0 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0] , [4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 2 lexStrict[ [0, 0, 0, 0, 4, 0, 4, 0, 0, 0, 0, 0, 0, 0] , [4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 6.1.3 Transition Removal

We remove transitions 6, 8 using the following ranking functions, which are bounded by −1.

 0: 0 1: 0 0_var_snapshot: −1 0*: 1
Hints:
 6 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 8 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 6.1.4 Splitting Cut-Point Transitions

We consider 1 subproblems corresponding to sets of cut-point transitions as follows.

### 6.1.4.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 5.

### 6.1.4.1.1 Splitting Cut-Point Transitions

There remain no cut-point transition to consider. Hence the cooperation termination is trivial.

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