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Proof step 10 from proof is-encoded-emptypage-middle (previous step)

# buf0 = EB_PAGE_SIZE, buf + emptypage + buf1 = buf0 + buf2, is-aligned(# buf, EB_PAGE_SIZE), is-encoded(nx0, buf2), is-encoded(rnode(m), buf0), EB_PAGE_SIZE ≤ # buf → (∃ buf0, buf1. buf = buf0 + buf1 ∧ # buf0 = EB_PAGE_SIZE), ∀ buf3, buf4. ¬ is-aligned(# buf3, EB_PAGE_SIZE) ∨ ¬ is-encoded(nx0, buf3 + emptypage + buf4) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# buf0 = EB_PAGE_SIZE, buf + emptypage + buf1 = buf0 + buf2, is-aligned(# buf, EB_PAGE_SIZE), is-encoded(rnode(m), buf0), is-encoded(nx0, buf2), # buf ≠ 0 → buf = buf3 + buf4 ∧ # buf3 = EB_PAGE_SIZE, ∀ buf3, buf4. ¬ is-aligned(# buf3, EB_PAGE_SIZE) ∨ ¬ is-encoded(nx0, buf3 + emptypage + buf4) ⊦

Rule argument

Additional information

Used simplifier rules: m ≠ 0 ⊦ is-aligned(n, m) → (m ≤ n ↔ n ≠ 0); (s)% ⊦ EB_PAGE_SIZE ≠ 0; (s)