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Proof step 10 from proof abs-node-subarray-failed (previous step)

# buf0 = NODE_HEADER_SIZE, # buf1 = nd.alignedsize, n = n0 + NODE_HEADER_SIZE, n < 2 * NODE_HEADER_SIZE + nd.alignedsize, is-aligned(n, NODE_HEADER_SIZE), is-encoded(nd, subarray(buf1, 0, node-size-headerless(nd))), is-encoded(nodeheader(node-size-headerless(nd), false), buf0), NODE_HEADER_SIZE ≤ n ⊦ n = 0, ∃ buf2, buf3. buf0 + subarray(buf1 + validtrailer, 0, n0) = buf2 + buf3 ∧ # buf2 = NODE_HEADER_SIZE ∧ (is-garbage-header(buf2) ∨ (∃ ndhd. is-encoded(ndhd, buf2) ∧ # buf3 ≤ align↑(ndhd.size, 2 * NODE_HEADER_SIZE)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m0 * 1 = m0; (s)%m1 ≠ 0 ⊦ m + n < (* m1) * m + n0 ↔ n < (m1 - 1) * m + n0; (s)%m ≠ 0, is-aligned(n1, m) ⊦ is-aligned(n0 + n1, m) ↔ is-aligned(n0, m); (s)%n ≠ 0 ⊦ is-aligned(n, n); (s)%m ≠ 0 ⊦ is-aligned(n, m) → (m ≤ n ↔ n ≠ 0); (s)% ⊦ NODE_HEADER_SIZE ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ NODE_HEADER_SIZE + n ≠ 0; (s)