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Proof step 66 from proof refine-add (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 67:
((del-list(to_delete0, n1) * ls(rys, ys) * ls(rxs, xs)))(H1), ((del-list(to_delete, n1 + 2) * ls(rys, ys) * ls(rxs, xs)))(H), ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n1 + 2)))(H), # xs + 1 < h0, # xs < h0, 2 ≤ # xs + # ys, (n1 + 2) + # xs ≤ h0, n1 + 2 ≤ # xs + # ys, ys = [] ∨ # ys = h0 ⊦ # xs + 1 = h0, # xs + # ys = 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: ⊦ m ≤ m + n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%n < n0 ⊦ (* n) ≠ m0 + (* n0); (s)