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Proof step 8 from proof abs-nodes-padding-prefix (previous step)

abs-padding(buf0) ⊦ abs-nodes(buf0 + buf1, m, nd ' + x0, lax, weak) ↔ abs-nodes(buf1, m + # buf0, nd ' + x0, lax, weak)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ abs-nodes(buf, m, a ' + x, lax, weak) ↔ (∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, a ' + x, lax, weak) ∨ lax ≠ [] ∧ lax.head.off = m ∧ lax.head.size = # buf0 ∧ abs-node(buf0, a) ∧ abs-nodes(buf1, m + # buf0, x, lax.tail, weak)))
Proof step 10:
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)

Rule argument

spec: <toplevel> name: abs-nodes-rec seq: ⊦ abs-nodes(buf, m, a ' + x, lax, weak) ↔ (∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, a ' + x, lax, weak) ∨ lax ≠ [] ∧ lax.head.off = m ∧ lax.head.size = # buf0 ∧ abs-node(buf0, a) ∧ abs-nodes(buf1, m + # buf0, x, lax.tail, weak))) subst: [weak, lax, x, a, m, buf] → [weak, lax, x0, nd, m, buf0 + buf1] flags: <all paths>

Additional information

Used terms: weak, lax, x0, nd, m, buf0 + buf1