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Proof step 11 from proof abs-nodes-one-h (previous step)

⊦ (∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, a ', ladr ' + lax0, weak) ∨ ladr.off = m ∧ ladr.size = # buf0 ∧ abs-node(buf0, a) ∧ abs-nodes(buf1, ladr.off + # buf0, [], lax0, weak))) ↔ (∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, a ', ladr ' + lax0, weak) ∨ ladr ' + lax0 = @(m, 2 * NODE_HEADER_SIZE + a.alignedsize) ' ∧ abs-node(buf0, a) ∧ abs-nodes(buf1, m + # buf0, [], [], weak)))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
ladr = @(m0, m1) ⊦ (∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, a ', ladr ' + lax0, weak) ∨ m0 = m ∧ m1 = # buf0 ∧ abs-node(buf0, a) ∧ abs-nodes(buf1, m0 + # buf0, [], lax0, weak))) ↔ (∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, a ', ladr ' + lax0, weak) ∨ ladr ' + lax0 = @(m, 2 * NODE_HEADER_SIZE + a.alignedsize) ' ∧ abs-node(buf0, a) ∧ abs-nodes(buf1, m + # buf0, [], [], weak)))

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