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Proof step 32 from proof NEWPARTS-example (previous step)

Pdone = pe(0, 0, 0) ' + pe(3, 4, 0) ', Pnew = pe(1, 2, 0) ' + pe(5, 5, 0) ', F0 = (∅)[0, 0][1, 0][2, 0][3, 1][4, 1][5, 1][1, 2, 2][5, 5, 3], P0 = [], length_of_P = 2 ⊦ ⟨while P0 ≠ [] do let p = P0.head in {P0 := P0.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) ', F0 := F0[p.first + p.count, p.last, length_of_P + # Pnew] else Pdone := Pdone + p '}⟩ ⟨P0 := Pdone + Pnew⟩ ( P0 = pe(0, 0, 0) ' + pe(3, 4, 0) ' + pe(1, 2, 0) ' + pe(5, 5, 0) ' ∧ F0 = (∅)[0, 0][1, 2][2, 2][3, 1][4, 1][5, 3])

Heuristic loop exit applied rule while exit right

And obtained the following new goals

Proof step 31:
Pdone = pe(0, 0, 0) ' + pe(3, 4, 0) ', Pnew = pe(1, 2, 0) ' + pe(5, 5, 0) ', F0 = (∅)[0, 0][1, 0][2, 0][3, 1][4, 1][5, 1][1, 2, 2][5, 5, 3], P0 = [], length_of_P = 2 ⊦ ⟨P0 := Pdone + Pnew⟩ ( P0 = pe(0, 0, 0) ' + pe(3, 4, 0) ' + pe(1, 2, 0) ' + pe(5, 5, 0) ' ∧ F0 = (∅)[0, 0][1, 2][2, 2][3, 1][4, 1][5, 3])

Rule argument

Additional information

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