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Proof step 16 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, @(m0, m1) ' + lax0, weak) ∨ m0 = m ∧ m1 = # buf ∧ abs-node(buf, nd) ∧ abs-nodes(buf2, m + # buf, x0, lax0, weak)) ⊦ abs-nodes(buf1, m + # buf0, nd ' + x0, @(m0, m1) ' + lax0, weak)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
buf0 + buf1 = buf2 + buf, abs-padding(buf0), buf0 = buf2 ∧ abs-nodes(buf, m + # buf2, nd ' + x0, @(m0, m1) ' + lax0, weak) ∨ m0 = m ∧ m1 = # buf2 ∧ abs-node(buf2, nd) ∧ abs-nodes(buf, m + # buf2, x0, lax0, weak) ⊦ abs-nodes(buf1, m + # buf0, nd ' + x0, @(m0, m1) ' + lax0, weak)

Proof step 16 from proof abs-nodes-padding-prefix (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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