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Proof step 5 from proof abs-nodes-aligned-2headersize-h (previous step)

x = [], abs-nodes(buf, m, x, lax, false), ∀ buf0, m, x, lax. # buf0 < # buf → abs-nodes(buf0, m, x, lax, false) → is-aligned(# buf0, 2 * NODE_HEADER_SIZE) ⊦ is-aligned(# buf, 2 * NODE_HEADER_SIZE)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
abs-nodes(buf, m, [], [], false), ∀ buf0, m, x, lax. # buf0 < # buf → abs-nodes(buf0, m, x, lax, false) → is-aligned(# buf0, 2 * NODE_HEADER_SIZE) ⊦ is-aligned(# buf, 2 * NODE_HEADER_SIZE)

Rule argument

Additional information

Used simplifier rules: abs-nodes-elems-empty-adr-empty-h: ⊦ abs-nodes(buf, m, [], lax, weak) → lax = [];