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Proof step 13 from proof abs-nodes-padding-prefix (previous step)

abs-padding(buf0) ⊦ (∃ buf, buf2. buf0 + buf1 = buf + buf2 ∧ ( buf0 = buf ∧ abs-nodes(buf2, m + # buf, nd ' + x0, ladr ' + lax0, weak) ∨ ladr.off = m ∧ ladr.size = # buf ∧ abs-node(buf, nd) ∧ abs-nodes(buf2, ladr.off + # buf, x0, lax0, weak))) ↔ abs-nodes(buf1, m + # buf0, nd ' + x0, ladr ' + lax0, weak)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
ladr = @(m0, m1), abs-padding(buf0) ⊦ (∃ buf, buf2. buf0 + buf1 = buf + buf2 ∧ ( buf0 = buf ∧ abs-nodes(buf2, m + # buf, nd ' + x0, ladr ' + lax0, weak) ∨ m0 = m ∧ m1 = # buf ∧ abs-node(buf, nd) ∧ abs-nodes(buf2, m0 + # buf, x0, lax0, weak))) ↔ abs-nodes(buf1, m + # buf0, nd ' + x0, ladr ' + lax0, 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