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Proof step 15 from proof abs-nodes-at-least-flashsize (previous step)

lax.head = @(m0, m1), m0 = m, m1 = # buf1, # buf1 + # buf0 < (2 * NODE_HEADER_SIZE + nd.alignedsize) + x0.flashsize, abs-node(buf1, nd), abs-nodes(buf0, m + # buf1, x0, lax.tail, false), ∀ buf2, x, m, lax. # buf2 < #(buf1 + buf0) → # buf2 < x.flashsize ∧ abs-nodes(buf2, m, x, lax, false) → x = [] ⊦ lax = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# buf1 = nd.alignedsize + 2 * NODE_HEADER_SIZE, lax.head = @(m, # buf1), # buf0 < x0.flashsize, abs-node(buf1, nd), abs-nodes(buf0, m + # buf1, x0, lax.tail, false), ∀ buf2, x, m, lax. # buf2 < #(buf1 + buf0) → # buf2 < x.flashsize ∧ abs-nodes(buf2, m, x, lax, false) → x = [] ⊦ lax = [], x0 = []

Rule argument

Additional information

Used simplifier rules: ls-abs-node-buf-size: ⊦ abs-node(buf, nd) → # buf = nd.alignedsize + 2 * NODE_HEADER_SIZE;%flashsize-eq-0: ⊦ x.flashsize = 0 ↔ x = [];%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic