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Proof step 20 from proof abs-nodes-unique-strong-h (previous step)

# buf1 = nd.alignedsize + 2 * NODE_HEADER_SIZE, lax0.head = @(m, # buf1), abs-node(buf1, nd), abs-nodes(buf0, m + # buf1, x, lax0.tail, false), ∀ buf2, m, x0, lax0, x1, lax1, boolvar1. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x0, lax0, false) → (abs-nodes(buf2, m, x1, lax1, boolvar1) ↔ x0 = x1 ∧ lax0 = lax1) ⊦ abs-nodes(buf1 + buf0, m, x1, lax1, boolvar1) ↔ nd ' + x = x1 ∧ lax0 = lax1, lax0 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
# buf1 = nd.alignedsize + 2 * NODE_HEADER_SIZE, ladr = @(m, # buf1), lax0 = ladr + lax2, abs-node(buf1, nd), abs-nodes(buf0, m + # buf1, x, lax2, false), ∀ buf2, m, x0, lax0, x1, lax1, boolvar1. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x0, lax0, false) → (abs-nodes(buf2, m, x1, lax1, boolvar1) ↔ x0 = x1 ∧ lax0 = lax1) ⊦ abs-nodes(buf1 + buf0, m, x1, lax1, boolvar1) ↔ nd ' + x = x1 ∧ lax0 = lax1, lax0 = []

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] → [lax0, ladr, lax]

Additional information

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