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Proof step 16 from proof abs-nodes-aligned-2headersize-h (previous step)

abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x0, lax, false) ∨ lax ≠ [] ∧ lax.head.off = m ∧ lax.head.size = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, lax.head.off + # buf1, x0, lax.tail, false), ∀ buf2, m, x, lax. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, false) → is-aligned(# buf2, 2 * NODE_HEADER_SIZE) ⊦ is-aligned(# buf1 + # buf0, 2 * NODE_HEADER_SIZE)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
lax.head = @(m0, m1), abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x0, lax, false) ∨ lax ≠ [] ∧ m0 = m ∧ m1 = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, m0 + # buf1, x0, lax.tail, false), ∀ buf2, m, x, lax. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, false) → is-aligned(# buf2, 2 * NODE_HEADER_SIZE) ⊦ is-aligned(# buf1 + # buf0, 2 * NODE_HEADER_SIZE)

Rule argument

spec: laddress name: elim seq: ⊦ m = ladr.off ∧ m0 = ladr.size ↔ ladr = @(m, m0) subst: [ladr, m, m0] → [lax.head, m0, m1]

Additional information

Used simplifier rules: ⊦ m = ladr.off ∧ m0 = ladr.size ↔ ladr = @(m, m0); (elim) Used terms: lax.head, m0, m1