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Proof step 6 from proof cleanup-lpt-n-read-above (previous step)

lpt[n0] = mklp(m0, m1, lpf, m2), n0 + 1 < # lpt, m < # lpt, n0 < m, n0 < # lpt, ∀ lpt, m. n0 < # lpt ∧ m < # lpt ∧ n0 ≤ m → cleanup-lpt(lpt, n0)[m] = lpt[m] ⊦ (m0 = 0 ∧ lpf ≠ LP_FREE ⊃ cleanup-lpt(lpt, n0)[n0, mklp(0, 0, LP_FREE, m2)];cleanup-lpt(lpt, n0))[m] = lpt[m]

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 7:
lpt[n0] = mklp(m0, m1, lpf, m2), m0 = 0, n0 < # lpt, n0 < m, m < # lpt, n0 + 1 < # lpt, ∀ lpt, m. n0 < # lpt ∧ m < # lpt ∧ n0 ≤ m → cleanup-lpt(lpt, n0)[m] = lpt[m] ⊦ cleanup-lpt(lpt, n0)[n0, mklp(0, 0, LP_FREE, m2)][m] = lpt[m], lpf = LP_FREE
Proof step 9:
lpt[n0] = mklp(m0, m1, lpf, m2), n0 < # lpt, n0 < m, m < # lpt, n0 + 1 < # lpt, ∀ lpt, m. n0 < # lpt ∧ m < # lpt ∧ n0 ≤ m → cleanup-lpt(lpt, n0)[m] = lpt[m], m0 ≠ 0 ∨ lpf = LP_FREE ⊦ cleanup-lpt(lpt, n0)[m] = lpt[m]

Rule argument

pos: right 1

Proof step 6 from proof cleanup-lpt-n-read-above (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information