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Proof step 126 from proof inv-format (previous step)

# pebs0 = n0 + m1, # pebs1 = n0 + m1, pebs1[m1] = mkpeb(buf, m2, false), aehdr = aechdr((err1 = ESUCCESS ∧ aehdr0.valid? ⊃ aehdr0.ec + 1;0)), m = EB_PAGE_SIZE, m0 = m1, pebs = pebs1[m1, mkpeb(emptypeb, 0, false)], mtd-cons(pebs0), mtd-cons(pebs1), mtd-cons(pebs1[m1, mkpeb(emptypeb, 0, false)]), ubi-io-initial(subarray(pebs1, 0, m1)), m0 < # pebs → aehdr.valid? → m0 < # pebs ∧ (¬ pebs[m0].bad → m0 < # pebs), (err1 = ESUCCESS ∧ aehdr0.valid? ⊃ aehdr0.valid?;true) ⊦ n0 = 0, m0 < # pebs ∧ aehdr.valid? ∧ ¬ pebs[m0].bad ∧ pebs[m0].fill = 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ mkpeb(buf, m, boolvar).fill = m; (s)% ⊦ mkpeb(buf, m, boolvar).bad ↔ boolvar; (s)% ⊦ m < # pebs → pebs[m, peb][m] = peb; (s)% ⊦ aechdr(m).valid?; (s)% ⊦ # pebs[n, peb] = # pebs; (s)