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Proof step 123 from proof inv-format (previous step)

# pebs0 = n0 + m1, # pebs1 = n0 + m1, # buf1 = EB_PAGE_SIZE, pebs1[m1] = mkpeb(buf0, m2, false), pebs = pebs1[m1, mkpeb(emptypeb, 0, false)], m1 < # pebs1, mtd-cons(pebs0), mtd-cons(pebs1), mtd-cons(pebs1[m1, mkpeb(emptypeb, 0, false)]), ubi-io-initial(subarray(pebs1, 0, m1)), err = ESUCCESS → is-encoded(echdr((err1 = ESUCCESS ∧ aehdr0.valid? ⊃ aehdr0.ec + 1;0)), buf1), err = ESUCCESS ∨ ⌊err⌋ ⊦ ⟪m1 := m1 + 1⟫ ( m1 ≤ # pebs0 ∧ # pebs0 = # pebs ∧ mtd-cons(pebs) ∧ (err = ESUCCESS → ubi-io-initial(subarray(pebs, 0, m1))) ∧ (err = ESUCCESS ∨ ⌊err⌋ ) ∧ # pebs0 < n0 + m1), err = ESUCCESS, n0 = 0

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

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