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Proof step 18 from proof abs-nodes-aligned-2headersize-h (previous step)

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, m + # 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)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 2

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