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Proof step 21 from proof abs-nodes-subarray-failed (previous step)

# buf1 = nd.alignedsize + 2 * NODE_HEADER_SIZE, # buf0 = x0.flashsize, n < # buf1 + x0.flashsize, abs-node(buf1, nd), abs-nodes(buf0, m0 + # buf1, x0, lax0, false), is-aligned(n, NODE_HEADER_SIZE), # buf1 ≤ n, n ≤ # buf1 + # buf0, ∀ buf2, x, n, m, lax. # buf2 < #(buf1 + buf0) → # buf2 = x.flashsize ∧ n < x.flashsize ∧ abs-nodes(buf2, m, x, lax, false) ∧ is-aligned(n, NODE_HEADER_SIZE) → n = 0 ∨ x = [] ∨ (∃ n0. n0 < # lax ∧ abs-nodes(subarray(buf2, 0, n), m, prefixn(n0, x), prefixn(n0, lax), true)) ⊦ buf1 + buf0 = mkbuf(0), n = 0, ∃ n0. ¬ # lax0 < n0 ∧ abs-nodes(buf1 + subarray(buf0, 0, n - # buf1), m0, prefixn(n0, nd ' + x0), prefixn(n0, @(m0, # buf1) ' + lax0), true)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: flashsize-eq-0: ⊦ x.flashsize = 0 ↔ x = [];% ⊦ m < n → m ≤ n; (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 < m + n ↔ n ≠ 0; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat