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Proof step 159 from proof abs-nodes-add-crash-aligned-h (previous step)

# buf1 = a.alignedsize + 2 * NODE_HEADER_SIZE, lax.head = @(m0, m1), n = n0 + # buf2, n < (# buf2 + # buf) + # buf1, abs-node(buf1, a), abs-nodes(buf2 + subarray(buf + buf1, 0, n0), m, (a0 ' + x0) + a ', lax + @(m + # buf2 + # buf, # buf1) ', boolvar), is-aligned(m, NODE_HEADER_SIZE), is-aligned(n, NODE_HEADER_SIZE), # buf2 ≤ n, abs-padding(buf2) ∧ abs-nodes(buf, m + # buf2, a0 ' + x0, lax, false) ∨ lax ≠ [] ∧ m0 = m ∧ m1 = # buf2 ∧ abs-node(buf2, a0) ∧ abs-nodes(buf, m + # buf2, x0, lax.tail, false), ∀ buf3, buf1, a, n, m, x, lax, boolvar. # buf3 < #(buf2 + buf) → # buf1 = a.alignedsize + 2 * NODE_HEADER_SIZE ∧ n < # buf3 + # buf1 ∧ abs-node(buf1, a) ∧ abs-nodes(subarray(buf3 + buf1, 0, n), m, x + a ', lax + @(m + # buf3, # buf1) ', boolvar) ∧ abs-nodes(buf3, m, x, lax, false) ∧ is-aligned(n, NODE_HEADER_SIZE) ∧ is-aligned(m, NODE_HEADER_SIZE) → n = # buf3 + # buf1 ⊦ # buf + # buf1 = 0, n = (# buf2 + # buf) + # buf1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 160:
# buf1 = a.alignedsize + 2 * NODE_HEADER_SIZE, lax.head = @(m0, m1), n0 < # buf + # buf1, abs-node(buf1, a), abs-nodes(buf2 + subarray(buf + buf1, 0, n0), m, (a0 ' + x0) + a ', lax + @(m + # buf2 + # buf, # buf1) ', boolvar), is-aligned(m, NODE_HEADER_SIZE), is-aligned(n0 + # buf2, NODE_HEADER_SIZE), abs-padding(buf2) ∧ abs-nodes(buf, m + # buf2, a0 ' + x0, lax, false) ∨ lax ≠ [] ∧ m0 = m ∧ m1 = # buf2 ∧ abs-node(buf2, a0) ∧ abs-nodes(buf, m + # buf2, x0, lax.tail, false), ∀ buf3, buf1, a, n, m, x, lax, boolvar. # buf3 < #(buf2 + buf) → # buf1 = a.alignedsize + 2 * NODE_HEADER_SIZE ∧ n < # buf3 + # buf1 ∧ abs-node(buf1, a) ∧ abs-nodes(subarray(buf3 + buf1, 0, n), m, x + a ', lax + @(m + # buf3, # buf1) ', boolvar) ∧ abs-nodes(buf3, m, x, lax, false) ∧ is-aligned(n, NODE_HEADER_SIZE) ∧ is-aligned(m, NODE_HEADER_SIZE) → n = # buf3 + # buf1 ⊦ n0 = # buf + # buf1

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m ≤ m + n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)