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Proof step 14 from proof abs-padding-subarray-failed (previous step)

# buf0 = NODE_HEADER_SIZE, n0 < # buf1, n0 < NODE_HEADER_SIZE + # buf1, is-aligned(n0, NODE_HEADER_SIZE), is-aligned(# buf1, 2 * NODE_HEADER_SIZE), is-encoded(nodeheader(# buf1, true), buf0) ⊦ buf1 = mkbuf(0), ∃ buf, buf2. buf0 + subarray(buf1, 0, n0) = buf + buf2 ∧ # buf = NODE_HEADER_SIZE ∧ (is-garbage-header(buf) ∨ (∃ ndhd. is-encoded(ndhd, buf) ∧ # buf2 ≤ align↑(ndhd.size, 2 * NODE_HEADER_SIZE)))

User applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: buf0, subarray(buf1, 0, n0) abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ is-encoded(ndhd, buf) → ¬ is-garbage-header(buf); (s)% ⊦ is-encoded(ndhd, buf) → (is-encoded(ndhd0, buf) ↔ ndhd = ndhd0); (s)% ⊦ # subarray(buf, m, n) = n; (s)% ⊦ nodeheader(m, boolvar).size = m; (s)% ⊦ 2 * m = 0 ↔ m = 0; (s)% ⊦ NODE_HEADER_SIZE ≠ 0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%m ≠ 0 ⊦ is-aligned(n, m) → n % m = 0; (s)%n ≠ 0, m % n = 0 ⊦ align↑(m, n) = m; (s)% ⊦ m < n → m ≤ n; (s) Used terms: buf0, subarray(buf1, 0, n0)