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Proof step 42 from proof NEWPARTS-example (previous step)

P1 = pe(0, 2, 1) ' + pe(3, 5, 2) ', length_of_P = 2, F1 = (∅)[0, 0][1, 0][2, 0][3, 1][4, 1][5, 1], Pdone = [], Pnew = [], p = P1.head ⊦ ⟨P1 := P1.tail⟩ ⟨if p.count ≠ 0 ∧ p.first + p.count ≤ p.last then Pdone := Pdone + pe(p.first, p.first + (p.count - 1), 0) ', Pnew := Pnew + pe(p.first + p.count, p.last, 0) ', F1 := F1[p.first + p.count, p.last, length_of_P + # Pnew] else Pdone := Pdone + p '⟩ ⟨while P1 ≠ [] do let p = P1.head in {P1 := P1.tail; if p.count ≠ 0 ∧ p.first + p.count ≤ p.last then Pdone := Pdone + pe(p.first, p.first + (p.count - 1), 0) ', Pnew := Pnew + pe(p.first + p.count, p.last, 0) ', F1 := F1[p.first + p.count, p.last, length_of_P + # Pnew] else Pdone := Pdone + p '}⟩ ⟨P1 := Pdone + Pnew⟩ ( P1 = pe(0, 0, 0) ' + pe(3, 4, 0) ' + pe(1, 2, 0) ' + pe(5, 5, 0) ' ∧ F1 = (∅)[0, 0][1, 2][2, 2][3, 1][4, 1][5, 3])

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 41:
P1 = pe(0, 2, 1) ' + pe(3, 5, 2) ', length_of_P = 2, F1 = (∅)[0, 0][1, 0][2, 0][3, 1][4, 1][5, 1], Pdone = [], Pnew = [], p = P1.head, P = P1.tail ⊦ ⟨if p.count ≠ 0 ∧ p.first + p.count ≤ p.last then Pdone := Pdone + pe(p.first, p.first + (p.count - 1), 0) ', Pnew := Pnew + pe(p.first + p.count, p.last, 0) ', F1 := F1[p.first + p.count, p.last, length_of_P + # Pnew] else Pdone := Pdone + p '⟩ ⟨while P ≠ [] do let p = P.head in {P := P.tail; if p.count ≠ 0 ∧ p.first + p.count ≤ p.last then Pdone := Pdone + pe(p.first, p.first + (p.count - 1), 0) ', Pnew := Pnew + pe(p.first + p.count, p.last, 0) ', F1 := F1[p.first + p.count, p.last, length_of_P + # Pnew] else Pdone := Pdone + p '}⟩ ⟨P := Pdone + Pnew⟩ ( P = pe(0, 0, 0) ' + pe(3, 4, 0) ' + pe(1, 2, 0) ' + pe(5, 5, 0) ' ∧ F1 = (∅)[0, 0][1, 2][2, 2][3, 1][4, 1][5, 3])

Rule argument

Additional information

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