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Proof step 74 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) ', 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) ')

Heuristic symbolic execution applied rule if positive right

And closed the goal!

Rule argument

pos:right 1

Additional information

[Concretesimprules((Csimpnamedseq(in-insert-strong,,, ⊦ m0 ∈ D[m, n] ↔ ¬ ¬(m0 = m ∨ m0 ∈ D))))]