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Proof step 2 from proof refine-add (previous step)

# xs = xs_len0, # xs < h0, ∃ n. ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n)))(H) ∧ n ≤ # xs + # ys ∧ n + # xs ≤ h ∧ h = h0 ∧ xs_len = xs_len0, # ys = 0 ∨ # ys = h0 ⊦ ⟪tree_buffer_heap_add#(a, h; H, rxs, rys, xs_len, to_delete)⟫ ⟨caterpillar_add#(a, h0; xs, xs_len0, ys)⟩ ∃ n. ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n)))(H) ∧ n ≤ # xs + # ys ∧ n + # xs ≤ h ∧ h = h0 ∧ xs_len = xs_len0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
xs_len = # xs, ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n)))(H), # xs < h0, n + # xs ≤ h0, n ≤ # xs + # ys, ys = [] ∨ # ys = h0 ⊦ ⟪tree_buffer_heap_add#(a, h0; H, rxs, rys, xs_len, to_delete)⟫ ⟨xs_len0 := # xs; caterpillar_add#(a, h0; xs, xs_len0, ys)⟩ ∃ n. ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n)))(H) ∧ n ≤ # xs + # ys ∧ n + # xs ≤ h0 ∧ xs_len = xs_len0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)