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Proof step 25 from proof inv-format (previous step)

# pebs0 = # pebs, # pebs0 = n0 + m1, m1 < # pebs, mtd-cons(pebs0), mtd-cons(pebs), ubi-io-initial(subarray(pebs, 0, m1)), m1 ≤ # pebs0 ⊦ ⟪mtd_asm_isbad#(m1, pebs; ; isbad, err); m := m1; if err = ESUCCESS ∧ ¬ isbad then let aehdr = ?, bitflips = ? in {ubi_io_read_echdr#(m, EB_PAGE_SIZE, pebs; aehdr; bitflips, err); let n = (err = ESUCCESS ∧ aehdr.valid? ⊃ aehdr.ec + 1;0) in {mtd_asm_erase#(m; pebs; err); if err = ESUCCESS then ubi_io_write_echdr#(m, aechdr(n), EB_PAGE_SIZE; pebs; err)}}; m := m + 1⟫ ( m ≤ # pebs0 ∧ # pebs0 = # pebs ∧ mtd-cons(pebs) ∧ (err = ESUCCESS → ubi-io-initial(subarray(pebs, 0, m))) ∧ (err = ESUCCESS ∨ ⌊err⌋ ) ∧ # pebs0 < n0 + m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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