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Proof step 65 from proof refine-add (previous step)

((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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 66:
n = n1 + 2, ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n)))(H), ((del-list(to_delete, n) * ls(rys, ys) * ls(rxs, xs)))(H), ((del-list(to_delete0, n1) * ls(rys, ys) * ls(rxs, xs)))(H1), # xs < h0, # xs + 1 < h0, 2 ≤ n, n ≤ # xs + # ys, n + # xs ≤ h0, 2 ≤ # xs + # ys, ys = [] ∨ # ys = h0 ⊦ # xs + # ys = 0, # xs + 1 = h0, 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

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [n, m, n0] → [2, n, n1]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: 2, n, n1