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

abs-node(buf, nd), isempty(buf, 0, NODE_HEADER_SIZE) ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

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(ndhd, mkempbuf(n)); (s)% ⊦ # mkempbuf(n) = n; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ isempty(buf) ↔ buf = mkempbuf(# buf); (s)%m = # buf0 ⊦ isempty(buf0 + buf1, 0, m) ↔ isempty(buf0); (s)%is-encoded-size: ⊦ is-encoded(ndhd, buf) → # buf = NODE_HEADER_SIZE; (forward) from specification encoding-fixed-size-nonempty[encoding-node-header]