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Proof step 10 from proof abs-nodes-aligned-2headersize-h (previous step)

buf = buf0 + buf1, abs-nodes(buf1, m + # buf0, [], [], false), abs-padding(buf0), ∀ 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 11:
abs-nodes(buf1, m + # buf0, [], [], false), abs-padding(buf0), ∀ buf2, m, x, lax. # buf2 < #(buf0 + buf1) → abs-nodes(buf2, m, x, lax, false) → is-aligned(# buf2, 2 * NODE_HEADER_SIZE) ⊦ is-aligned(# buf1, 2 * NODE_HEADER_SIZE)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%m ≠ 0, is-aligned(n1, m) ⊦ is-aligned(n0 + n1, m) ↔ is-aligned(n0, m); (s)% ⊦ abs-padding(buf) → is-aligned(# buf, 2 * NODE_HEADER_SIZE); (ss)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ NODE_HEADER_SIZE ≠ 0; (s)% ⊦ 2 * m = 0 ↔ m = 0; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ #(buf + buf0) = # buf + # buf0; (s)