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Proof step 25 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) ⊦ ∃ buf4, buf3. subarray((buf1 + buf2 + validtrailer) + buf0, 0, 2 * NODE_HEADER_SIZE + align↑(node-size-headerless(nd1), 2 * NODE_HEADER_SIZE)) = buf4 + buf3 + validtrailer ∧ is-encoded(nodeheader(node-size-headerless(nd1), false), buf4) ∧ is-encoded(nd1, subarray(buf3, 0, node-size-headerless(nd1))) ∧ # buf3 = align↑(node-size-headerless(nd1), 2 * NODE_HEADER_SIZE)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ buf + buf0 = buf1 + buf0 ↔ buf = buf1; (s)%n = # buf ⊦ subarray(buf + buf0, 0, n) = buf; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m0 * 1 = m0; (s)%m2 ≠ 0 ⊦ (* m2) * m = m + n0 ↔ (m2 - 1) * m = n0; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ # validtrailer = NODE_HEADER_SIZE; (s)% ⊦ #(buf + buf0) = # buf + # buf0; (s)