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Proof step 26 from proof abs-node-abs-node-append-subarray (previous step)

# buf1 = NODE_HEADER_SIZE, # buf2 = align↑(node-size-headerless(nd1), 2 * NODE_HEADER_SIZE), is-encoded(nd1, subarray(buf2, 0, node-size-headerless(nd1))), is-encoded(nodeheader(node-size-headerless(nd1), false), buf1) ⊦ ∃ buf, buf3. buf1 + buf2 = buf + buf3 ∧ is-encoded(nodeheader(node-size-headerless(nd1), false), buf) ∧ is-encoded(nd1, subarray(buf3, 0, node-size-headerless(nd1))) ∧ # buf3 = # buf2

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: buf1, buf2

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: buf1, buf2