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Proof step 15 from proof abs-node-subarray-failed (previous step)

# buf1 = nd.alignedsize, # buf0 = NODE_HEADER_SIZE, n0 < NODE_HEADER_SIZE + nd.alignedsize, n0 < nd.alignedsize, is-aligned(n0, NODE_HEADER_SIZE), is-encoded(nodeheader(node-size-headerless(nd), false), buf0), is-encoded(nd, subarray(buf1, 0, node-size-headerless(nd))) ⊦ buf0 + subarray(buf1, 0, n0) = buf0 + subarray(buf1, 0, n0) ∧ # buf0 = NODE_HEADER_SIZE ∧ (is-garbage-header(buf0) ∨ (∃ ndhd. is-encoded(ndhd, buf0) ∧ # subarray(buf1, 0, n0) ≤ 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: lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m + n < m0 → m + n < m0 + n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ nodeheader(m, boolvar).size = m; (s)% ⊦ # subarray(buf, m, n) = n; (s)% ⊦ is-encoded(ndhd, buf) → (is-encoded(ndhd0, buf) ↔ ndhd = ndhd0); (s)% ⊦ is-encoded(ndhd, buf) → ¬ is-garbage-header(buf); (s)