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Proof step 28 from proof abs-nodes-aligned-headersize-h (previous step)

# buf1 = nd.alignedsize + 2 * NODE_HEADER_SIZE, lax.head = @(m, # buf1), abs-node(buf1, nd), abs-nodes(buf0, m + # buf1, x0, lax.tail, weak), ∀ buf2, m, x, lax, weak. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, weak) → is-aligned(# buf2, NODE_HEADER_SIZE) ⊦ lax = [], is-aligned(# buf0, NODE_HEADER_SIZE)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 29:
# buf1 = nd.alignedsize + 2 * NODE_HEADER_SIZE, ladr = @(m, # buf1), lax = ladr + lax0, abs-node(buf1, nd), abs-nodes(buf0, m + # buf1, x0, lax0, weak), ∀ buf2, m, x, lax, weak. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, weak) → is-aligned(# buf2, NODE_HEADER_SIZE) ⊦ lax = [], is-aligned(# buf0, NODE_HEADER_SIZE)

Rule argument

spec: list-data, inst: laddress-list name: elim-list-c seq: lax ≠ [] ⊦ ladr = lax.head ∧ lax0 = lax.tail ↔ lax = ladr + lax0 subst: [lax, ladr, lax0] → [lax, ladr, lax0]

Additional information

Used simplifier rules: lax ≠ [] ⊦ ladr = lax.head ∧ lax0 = lax.tail ↔ lax = ladr + lax0; (elim) Used terms: lax, ladr, lax0