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Proof step 9 from proof cleanup-lpt-n-at (previous step)

lpt[m] = mklp(m0, m1, lpf, m2), lpt[n0] = mklp(m3, m4, lpf0, m5), n0 < # lpt, m < # lpt, m ≤ n0, m < n0 → cleanup-lpt(lpt, n0)[m] = (m0 = 0 ∧ lpf ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, m2);mklp(m0, m1, lpf, m2)) ⊦ (m3 = 0 ∧ lpf0 ≠ LP_FREE ⊃ cleanup-lpt(lpt, n0)[n0, mklp(0, 0, LP_FREE, m5)];cleanup-lpt(lpt, n0))[m] = (m0 = 0 ∧ lpf ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, m2);mklp(m0, m1, lpf, m2))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 10:
lpt[n0] = mklp(m3, m4, lpf0, m5), lpt[m] = mklp(m0, m1, lpf, m2), m3 = 0, m < # lpt, n0 < # lpt, m ≤ n0, m < n0 → cleanup-lpt(lpt, n0)[m] = (m0 = 0 ∧ lpf ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, m2);mklp(m0, m1, lpf, m2)) ⊦ cleanup-lpt(lpt, n0)[n0, mklp(0, 0, LP_FREE, m5)][m] = (m0 = 0 ∧ lpf ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, m2);mklp(m0, m1, lpf, m2)), lpf0 = LP_FREE

Rule argument

pos: right 1

Additional information

Used simplifier rules: cleanup-lpt-n-read-above: n ≤ m, m < # lpt ⊦ cleanup-lpt(lpt, n)[m] = lpt[m];% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ n ≠ n0 → (n < n0 ↔ ¬ n0 < n); (ss)% ⊦ m < n → ¬ n < m; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ m ≤ m; (s)%⊦ mklp(m, m0, lpf, m1) = mklp(m2, m3, lpf0, m4) ↔ m = m2 ∧ m0 = m3 ∧ lpf = lpf0 ∧ m1 = m4; (ws)