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Proof step 2 from proof abs-node-full (previous step)

# buf1 = nd.alignedsize, is-encoded(nodeheader(node-size-headerless(nd), false), buf0), is-encoded(nd, subarray(buf1, 0, node-size-headerless(nd))) ⊦ abs-node(buf0 + buf1 + validtrailer, nd)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# 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

Rule argument

Additional information

Used simplifier rules: abs-node: ⊦ abs-node(buf, nd) ↔ (∃ buf0, buf1. buf = buf0 + buf1 + validtrailer ∧ is-encoded(nodeheader(node-size-headerless(nd), false), buf0) ∧ is-encoded(nd, subarray(buf1, 0, node-size-headerless(nd))) ∧ # buf1 = nd.alignedsize);%is-encoded-size: ⊦ is-encoded(ndhd, buf) → # buf = NODE_HEADER_SIZE; (forward) from specification encoding-fixed-size-nonempty[encoding-node-header]%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ buf + buf0 = buf1 + buf0 ↔ buf = buf1; (s)