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Proof step 22 from proof abs-nodes-elems-adr-length-h (previous step)

# buf1 = nd.alignedsize + 2 * NODE_HEADER_SIZE, abs-node(buf1, nd), abs-nodes(buf0, m + # buf1, x0, lax0, weak), ∀ buf2, m, x, lax, weak. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, weak) → # x = # lax ⊦ # x0 = # lax0

User applied rule apply induction

And closed the goal!

Rule argument

subst: [buf2, m, x, lax, weak] → [buf0, m + # buf1, x0, lax0, weak]

Additional information

Used simplifier rules: ⊦ #(buf + buf0) = # buf + # buf0; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)% ⊦ # buf = 0 ↔ buf = mkbuf(0); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ # mkbuf(n) = n; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)% ⊦ 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 Used terms: buf0, m + # buf1, x0, lax0, weak