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Proof step 1 from proof initialization (conclusion)

n > 0 ⊦ ⟪tree_buffer_init#(n; ; h0, xs0)⟫ ⟨caterpillar_init#(n; ; h, xs, xs_len, ys)⟩ (prefixn(h0, xs + ys) = prefixn(h0, xs0) ∧ #(xs + ys) ≤ # xs0 ∧ h0 = h)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n ≠ 0 ⊦ ⟪tree_buffer_init#(n; ; h0, xs0)⟫ ⟨caterpillar_init#(n; ; h, xs, xs_len, ys)⟩ (prefixn(h, xs + ys) = prefixn(h, xs0) ∧ # xs + # ys ≤ # xs0 ∧ h0 = h)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)% ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ m > n ↔ n < m; (s)