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Proof step 11 from proof abs-nodes-padding-prefix (previous step)

abs-padding(buf0) ⊦ (∃ buf, buf2. buf0 + buf1 = buf + buf2 ∧ ( buf0 = buf ∧ abs-nodes(buf2, m + # buf, nd ' + x0, lax, weak) ∨ lax ≠ [] ∧ lax.head.off = m ∧ lax.head.size = # buf ∧ abs-node(buf, nd) ∧ abs-nodes(buf2, lax.head.off + # buf, x0, lax.tail, weak))) ↔ abs-nodes(buf1, m + # buf0, nd ' + x0, lax, weak)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
lax = ladr + lax0, abs-padding(buf0) ⊦ (∃ buf, buf2. buf0 + buf1 = buf + buf2 ∧ ( buf0 = buf ∧ abs-nodes(buf2, m + # buf, nd ' + x0, lax, weak) ∨ lax ≠ [] ∧ ladr.off = m ∧ ladr.size = # buf ∧ abs-node(buf, nd) ∧ abs-nodes(buf2, ladr.off + # buf, x0, lax0, weak))) ↔ abs-nodes(buf1, m + # buf0, nd ' + x0, lax, weak), 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: ⊦ buf + buf0 = buf + buf1 ↔ buf0 = buf1; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]%lax ≠ [] ⊦ ladr = lax.head ∧ lax0 = lax.tail ↔ lax = ladr + lax0; (elim) Used terms: lax, ladr, lax0