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Proof step 12 from proof abs-nodes-nodups-h (previous step)

ladr ∈ lax0 ∨ dups(lax0), abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x0, ladr ' + lax0, boolvar) ∨ ladr.off = m ∧ ladr.size = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, ladr.off + # buf1, x0, lax0, boolvar), ∀ buf2, m, x, lax, boolvar. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, boolvar) ∧ dups(lax) → false ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
ladr = @(m0, m1), abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x0, ladr ' + lax0, boolvar) ∨ m0 = m ∧ m1 = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, m0 + # buf1, x0, lax0, boolvar), ladr ∈ lax0 ∨ dups(lax0), ∀ buf2, m, x, lax, boolvar. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, boolvar) ∧ dups(lax) → false ⊦

Rule argument

spec: laddress name: elim seq: ⊦ m = ladr.off ∧ m0 = ladr.size ↔ ladr = @(m, m0) subst: [ladr, m, m0] → [ladr, m0, m1]

Additional information

Used simplifier rules: ⊦ m = ladr.off ∧ m0 = ladr.size ↔ ladr = @(m, m0); (elim) Used terms: ladr, m0, m1