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Proof step 2 from proof abs-nodes-append-empty-h (previous step)

abs-nodes(buf0, m, x0, lax0, false) ⊦ abs-nodes(buf0 + buf1, m, x0, lax0, weak) ↔ abs-nodes(buf1, m + # buf0, [], [], weak)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information