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Proof step 136 from proof inv-format (previous step)

# pebs = n0 + m1, # pebs0 = n0 + m1, err = err0, err = ESUCCESS, m = m1, m1 < # pebs, mtd-cons(pebs), mtd-cons(pebs0), ubi-io-initial(subarray(pebs, 0, m1)), err = ESUCCESS → m1 < # pebs, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ n0 = 0, err = ESUCCESS ∧ ¬ pebs[m1].bad, m + 1 ≤ # pebs0 ∧ # pebs0 = # pebs ∧ mtd-cons(pebs) ∧ (err = ESUCCESS → ubi-io-initial(subarray(pebs, 0, m + 1))) ∧ (err = ESUCCESS ∨ ⌊err⌋ ) ∧ # pebs0 < n0 + m + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 137:
# pebs = n0 + m1, # pebs0 = n0 + m1, pebs[m1].bad, mtd-cons(pebs), mtd-cons(pebs0), ubi-io-initial(subarray(pebs, 0, m1)) ⊦ n0 = 0, ubi-io-initial(subarray(pebs, 0, m1 + 1))

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ ¬ ⌊ESUCCESS⌋ ; (s)