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Proof step 3 from proof abs-node-full (previous step)

# buf1 = nd.alignedsize, # buf0 = NODE_HEADER_SIZE, is-encoded(nd, subarray(buf1, 0, node-size-headerless(nd))), is-encoded(nodeheader(node-size-headerless(nd), false), buf0) ⊦ ∃ buf, buf2. buf0 + buf1 = buf + buf2 ∧ is-encoded(nodeheader(node-size-headerless(nd), false), buf) ∧ is-encoded(nd, subarray(buf2, 0, node-size-headerless(nd))) ∧ # buf2 = nd.alignedsize

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: buf0, buf1

Additional information

Used simplifier rules: ⊦ is-encoded(ndhd, buf) → (is-encoded(ndhd0, buf) ↔ ndhd = ndhd0); (s)% ⊦ is-encoded(nd, buf) → (is-encoded(nd0, buf) ↔ nd = nd0); (s) Used terms: buf0, buf1