[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 75 from proof REFINE-example (previous step)

D1 = (∅)[100, 2][11, 5][22, 0][33, 1][44, 4][55, 3], A1 = mkarray(6)[0, 22][1, 33][2, 100][3, 55][4, 44][5, 11], P1 = pe(0, 2, 0) ' + pe(3, 5, 0) ' ⊦ ⟨P := P1, D := D1, A := A1, F := (∅)[0, 0][1, 0][2, 0][3, 1][4, 1][5, 1], x := 11⟩ ⟨if x ∈ D then let i = D[x] in let p = P[F[i]] in if i < p.first + p.count then skip else {if i ≠ p.first + p.count then {A := A[i, A[p.first + p.count]][p.first + p.count, A[i]]; D := D[A[i], i]; D := D[A[p.first + p.count], p.first + p.count]}; P := P[F[i], p.count:= (p.count + 1)]}⟩ ⟨X0 := 22 ' + 44 '⟩ ⟨F0 := (∅)[0, 0][1, 0][2, 0][3, 1][4, 1][5, 1]; P1 := P, D1 := D, A1 := A; while X0 ≠ [] do {REFINEONE(X0.head, F0; A1, D1, P1); X0 := X0.tail}⟩ ( A1 = mkarray(6)[0, 22][1, 33][2, 100][3, 11][4, 44][5, 55] ∧ D1 = (∅)[100, 2][11, 3][22, 0][33, 1][44, 4][55, 5] ∧ P1 = pe(0, 2, 1) ' + pe(3, 5, 2) ')

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 74:
D1 = (∅)[100, 2][11, 5][22, 0][33, 1][44, 4][55, 3], A1 = mkarray(6)[0, 22][1, 33][2, 100][3, 55][4, 44][5, 11], P1 = pe(0, 2, 0) ' + pe(3, 5, 0) ', P0 = P1 ∧ D0 = D1 ∧ A0 = A1 ∧ F1 = (∅)[0, 0][1, 0][2, 0][3, 1][4, 1][5, 1] ∧ x0 = 11 ⊦ ⟨if x0 ∈ D0 then let i = D0[x0] in let p = P0[F1[i]] in if i < p.first + p.count then skip else {if i ≠ p.first + p.count then {A0 := A0[i, A0[p.first + p.count]][p.first + p.count, A0[i]]; D0 := D0[A0[i], i]; D0 := D0[A0[p.first + p.count], p.first + p.count]}; P0 := P0[F1[i], p.count:= (p.count + 1)]}⟩ ⟨X0 := 22 ' + 44 '⟩ ⟨F0 := (∅)[0, 0][1, 0][2, 0][3, 1][4, 1][5, 1]; P1 := P0, D1 := D0, A1 := A0; while X0 ≠ [] do {REFINEONE(X0.head, F0; A1, D1, P1); X0 := X0.tail}⟩ ( A1 = mkarray(6)[0, 22][1, 33][2, 100][3, 11][4, 44][5, 55] ∧ D1 = (∅)[100, 2][11, 3][22, 0][33, 1][44, 4][55, 5] ∧ P1 = pe(0, 2, 1) ' + pe(3, 5, 2) ')

Rule argument

Additional information

[Concretesimprules(())]