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Proof step 9 from proof sorted-shift-left (previous step)

m0 < n, m0 < # zbrar, m0 + 1 < nused, m < # zbrar, m + 1 < # zbrar, m0 < m, m < nused, n < # zbrar, nused < # zbrar, m0 < nused, 2 + m0 < # zbrar, m0 + 1 < # zbrar, m < n, sorted(zbrar, 0, nused + 1), n ≤ nused ⊦ zbrar[m0].key < zbrar[m].key

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
m0 < n, m0 < # zbrar, m < # zbrar, m + 1 < # zbrar, m0 < m, m < nused, n < # zbrar, nused < # zbrar, m0 < nused, 2 + m0 < # zbrar, m0 + 1 < # zbrar, m < n, m0 + 1 < n, m0 + 1 < nused, sorted(zbrar, 0, nused + 1), n ≤ nused ⊦ zbrar[m0].key < zbrar[m].key

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%transitivity: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification oelem[nat-set]%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic