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Proof step 52 from proof inv-io-read (previous step)

# buf = # buf2, # buf1 = # buf2, # buf2 = n3 + n0 + n, UBI_READ_RETRIES = n1 + m4, pebs[m3] = mkpeb(buf0, m2, false), m3 < # pebs, mtd-cons(pebs), range=(buf, n0 + n, buf2, n0 + n, n3), range=(buf, 0, buf2, 0, n0), range=(buf1, n0 + n, buf, n0 + n, n3), range=(buf1, 0, buf, 0, n0), m4 ≤ UBI_READ_RETRIES, m4 ≤ UBI_READ_RETRIES + 1, n0 + n ≤ # buf2, m1 + n ≤ PEB_SIZE, ⌊err1⌋ , ⌊err0⌋ ⊦ ⟪m := m4 + 1⟫ (m ≤ UBI_READ_RETRIES + 1 ∧ UBI_READ_RETRIES < n1 + m)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ n ≤ n0 ∧ n1 ≤ m → n + n1 ≤ n0 + m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ m < n → ¬ n < m; (s)