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Proof step 10 from proof abs-nodes-beyond-no-address-h (previous step)

@(m0, m1) ∈ lax, (# buf1 + # buf0) + m ≤ m0, abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x0, lax, boolvar) ∨ lax ≠ [] ∧ lax.head.off = m ∧ lax.head.size = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, lax.head.off + # buf1, x0, lax.tail, boolvar), ∀ buf2, m, x, lax, boolvar, m0, m1. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, boolvar) ∧ @(m0, m1) ∈ lax ∧ # buf2 + m ≤ m0 → false ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
lax = ladr + lax0, @(m0, m1) ∈ lax, (# buf1 + # buf0) + m ≤ m0, abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x0, lax, boolvar) ∨ lax ≠ [] ∧ ladr.off = m ∧ ladr.size = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, ladr.off + # buf1, x0, lax0, boolvar), ∀ buf2, m, x, lax, boolvar, m0, m1. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, boolvar) ∧ @(m0, m1) ∈ lax ∧ # buf2 + m ≤ m0 → false ⊦ lax = []

Rule argument

spec: list-data, inst: laddress-list name: elim-list-c seq: lax ≠ [] ⊦ ladr = lax.head ∧ lax0 = lax.tail ↔ lax = ladr + lax0 subst: [lax, ladr, lax0] → [lax, ladr, lax0]

Additional information

Used simplifier rules: ⊦ ¬ ladr ∈ []; (s)%lax ≠ [] ⊦ ladr = lax.head ∧ lax0 = lax.tail ↔ lax = ladr + lax0; (elim) Used terms: lax, ladr, lax0