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Proof step 37 from proof NEWPARTS-example (previous step)

length_of_P = 2, P = pe(3, 5, 2) ', Pdone0 = pe(0, 0, 0) ', Pnew0 = pe(1, 2, 0) ', F = (∅)[0, 0][1, 0][2, 0][3, 1][4, 1][5, 1][1, 2, 2] ⊦ ⟨let p = P.head in {P := P.tail; if p.count ≠ 0 ∧ p.first + p.count ≤ p.last then Pdone0 := Pdone0 + pe(p.first, p.first + (p.count - 1), 0) ', Pnew0 := Pnew0 + pe(p.first + p.count, p.last, 0) ', F := F[p.first + p.count, p.last, length_of_P + # Pnew0] else Pdone0 := Pdone0 + p '}⟩ ⟨while P ≠ [] do let p = P.head in {P := P.tail; if p.count ≠ 0 ∧ p.first + p.count ≤ p.last then Pdone0 := Pdone0 + pe(p.first, p.first + (p.count - 1), 0) ', Pnew0 := Pnew0 + pe(p.first + p.count, p.last, 0) ', F := F[p.first + p.count, p.last, length_of_P + # Pnew0] else Pdone0 := Pdone0 + p '}⟩ ⟨P := Pdone0 + Pnew0⟩ ( P = pe(0, 0, 0) ' + pe(3, 4, 0) ' + pe(1, 2, 0) ' + pe(5, 5, 0) ' ∧ F = (∅)[0, 0][1, 2][2, 2][3, 1][4, 1][5, 3])

Heuristic symbolic execution applied rule vardecls right

And obtained the following new goals

Proof step 36:
length_of_P = 2, P = pe(3, 5, 2) ', Pdone0 = pe(0, 0, 0) ', Pnew0 = pe(1, 2, 0) ', F = (∅)[0, 0][1, 0][2, 0][3, 1][4, 1][5, 1][1, 2, 2], p = P.head ⊦ ⟨P := P.tail⟩ ⟨if p.count ≠ 0 ∧ p.first + p.count ≤ p.last then Pdone0 := Pdone0 + pe(p.first, p.first + (p.count - 1), 0) ', Pnew0 := Pnew0 + pe(p.first + p.count, p.last, 0) ', F := F[p.first + p.count, p.last, length_of_P + # Pnew0] else Pdone0 := Pdone0 + p '⟩ ⟨while P ≠ [] do let p = P.head in {P := P.tail; if p.count ≠ 0 ∧ p.first + p.count ≤ p.last then Pdone0 := Pdone0 + pe(p.first, p.first + (p.count - 1), 0) ', Pnew0 := Pnew0 + pe(p.first + p.count, p.last, 0) ', F := F[p.first + p.count, p.last, length_of_P + # Pnew0] else Pdone0 := Pdone0 + p '}⟩ ⟨P := Pdone0 + Pnew0⟩ ( P = pe(0, 0, 0) ' + pe(3, 4, 0) ' + pe(1, 2, 0) ' + pe(5, 5, 0) ' ∧ F = (∅)[0, 0][1, 2][2, 2][3, 1][4, 1][5, 3])

Rule argument

Additional information

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