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Proof step 16 from proof abs-node-subarray-failed (previous step)

# buf0 = NODE_HEADER_SIZE, # buf1 = nd.alignedsize, align↑(node-size-headerless(nd), 2 * NODE_HEADER_SIZE) + 1 < nd.alignedsize, n0 < nd.alignedsize, n0 < NODE_HEADER_SIZE + nd.alignedsize, align↑(node-size-headerless(nd), 2 * NODE_HEADER_SIZE) < n0, align↑(node-size-headerless(nd), 2 * NODE_HEADER_SIZE) < nd.alignedsize, align↑(node-size-headerless(nd), 2 * NODE_HEADER_SIZE) < NODE_HEADER_SIZE + nd.alignedsize, is-aligned(n0, NODE_HEADER_SIZE), is-encoded(nd, subarray(buf1, 0, node-size-headerless(nd))), is-encoded(nodeheader(node-size-headerless(nd), false), buf0) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: encoding-node-base name: alignedsize seq: ⊦ nd.alignedsize = align↑(node-size-headerless(nd), 2 * NODE_HEADER_SIZE) subst: [nd] → [nd] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 2 * m = 0 ↔ m = 0; (s)% ⊦ NODE_HEADER_SIZE ≠ 0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n ≠ 0, m % n = 0 ⊦ align↑(m, n) = m; (s)%# buf = n ⊦ subarray(buf, 0, n) = buf; (s)% ⊦ m < n → ¬ n < m; (s) Used terms: nd