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

abs-node(mkbuf(0), nd) ⊦

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);%# buf = n ⊦ subarray(buf, 0, n) = buf; (s)% ⊦ validtrailer ≠ mkbuf(0); (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ buf + buf0 = mkbuf(0) ↔ buf = mkbuf(0) ∧ buf0 = mkbuf(0); (ws)