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Proof step 132 from proof inv-format (previous step)

# pebs = n0 + m1, # pebs0 = n0 + m1, pebs[m1] = mkpeb(buf, m0, false), err = err0, err2 = err, mtd-cons(pebs), mtd-cons(pebs0), ubi-io-initial(subarray(pebs, 0, m1)), ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪m1 := m1 + 1⟫ ( m1 ≤ # pebs0 ∧ # pebs0 = # pebs ∧ mtd-cons(pebs) ∧ (err2 = ESUCCESS → ubi-io-initial(subarray(pebs, 0, m1))) ∧ (err2 = ESUCCESS ∨ ⌊err2⌋ ) ∧ # 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% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s)