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Proof step 10 from proof abs-nodes-padding-prefix (previous step)

abs-padding(buf0) ⊦ (∃ buf2, buf3. buf0 + buf1 = buf2 + buf3 ∧ ( abs-padding(buf2) ∧ abs-nodes(buf3, m + # buf2, nd ' + x0, lax, weak) ∨ lax ≠ [] ∧ lax.head.off = m ∧ lax.head.size = # buf2 ∧ abs-node(buf2, nd) ∧ abs-nodes(buf3, m + # buf2, x0, lax.tail, weak))) ↔ abs-nodes(buf1, m + # buf0, nd ' + x0, lax, weak)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
abs-padding(buf0) ⊦ (∃ buf, buf2. buf0 + buf1 = buf + buf2 ∧ ( buf0 = buf ∧ abs-nodes(buf2, m + # buf, nd ' + x0, lax, weak) ∨ lax ≠ [] ∧ lax.head.off = m ∧ lax.head.size = # buf ∧ abs-node(buf, nd) ∧ abs-nodes(buf2, lax.head.off + # buf, x0, lax.tail, weak))) ↔ abs-nodes(buf1, m + # buf0, nd ' + x0, lax, weak)

Rule argument

Additional information

Used simplifier rules: ⊦ buf0 + buf1 = buf2 + buf3 ∧ abs-padding(buf0) → (abs-padding(buf2) ↔ buf0 = buf2); (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]