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Proof step 11 from proof abs-nodes-no-padding-one (previous step)

# buf0 = a.alignedsize + 2 * NODE_HEADER_SIZE, ladr.size = # buf0, abs-node(buf0, a) ⊦ ladr = @(ladr.off, # buf0)

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

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