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Proof step 21 from proof abs-nodes-aligned-2headersize-h (previous step)

lax.head = @(m0, m1), 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) ⊦ lax = [], is-aligned(# buf1 + # buf0, 2 * NODE_HEADER_SIZE)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
# buf1 = nd.alignedsize + 2 * NODE_HEADER_SIZE, lax.head = @(m, # 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) ⊦ lax = [], is-aligned(# buf0, 2 * NODE_HEADER_SIZE)

Rule argument

Additional information

Used simplifier rules: ls-abs-node-buf-size: ⊦ abs-node(buf, nd) → # buf = nd.alignedsize + 2 * NODE_HEADER_SIZE;%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-node(buf, nd) → 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)