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Proof step 30 from proof abs-padding-abs-nodes-h (previous step)

# buf1 = NODE_HEADER_SIZE, # buf2 = nd.alignedsize + 2 * NODE_HEADER_SIZE, buf1 + buf0 + validtrailer = buf2 + buf, lax.head = @(m, # buf2), abs-node(buf2, nd), abs-nodes(buf, m + # buf2, x0, lax.tail, weak), is-aligned(# buf0, 2 * NODE_HEADER_SIZE), is-encoded(nodeheader(# buf0, true), buf1) ⊦ lax = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 31:
# buf2 = nd.alignedsize + 2 * NODE_HEADER_SIZE, # buf1 = NODE_HEADER_SIZE, buf1 + buf0 + validtrailer = buf2 + buf, ladr = @(m, # buf2), lax = ladr + lax0, abs-node(buf2, nd), abs-nodes(buf, m + # buf2, x0, lax0, weak), is-aligned(# buf0, 2 * NODE_HEADER_SIZE), is-encoded(nodeheader(# buf0, true), buf1) ⊦ lax = []

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