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Proof step 64 from proof refine-add (previous step)

((del-list(to_delete, n) * ls(rys, ys) * ls(rxs, xs)))(H), ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n)))(H), # xs + 1 < h0, # xs < h0, n + # xs ≤ h0, n ≤ # xs + # ys, 2 ≤ n, 2 ≤ n → ((del-list(to_delete0, n - 2) * ls(rys, ys) * ls(rxs, xs)))(H1), n < 2 → to_delete0 = [] ∧ ((ls(rys, ys) * ls(rxs, xs)))(H1), ys = [] ∨ # ys = h0 ⊦ # xs + 1 = h0, r = null, r ∈ H1, ∃ n0. ((ls(r, a ' + xs) * ls(rys, ys) * del-list(to_delete0, n0)))((H1 ++ r)[r, a × rxs]) ∧ n0 ≤ (# xs + 1) + # ys ∧ n0 + # xs < h0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 65:
((del-list(to_delete0, n - 2) * ls(rys, ys) * ls(rxs, xs)))(H1), ((del-list(to_delete, n) * ls(rys, ys) * ls(rxs, xs)))(H), ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n)))(H), # xs + 1 < h0, # xs < h0, 2 ≤ # xs + # ys, n + # xs ≤ h0, n ≤ # xs + # ys, 2 ≤ n, ys = [] ∨ # ys = h0 ⊦ # xs + 1 = h0, # xs + # ys = 0, n = 0, r = null, r ∈ H1, ∃ n0. ((ls(r, a ' + xs) * ls(rys, ys) * del-list(to_delete0, n0)))((H1 ++ r)[r, a × rxs]) ∧ n0 ≤ (# xs + 1) + # ys ∧ n0 + # xs < h0

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat% ⊦ m ≤ n → (n < m ↔ false); (s)