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Proof step 8 from proof keys-shift-left (previous step)

nused < # zbrar, n < # zbrar, sorted(zbrar, 0, nused + 1), n ≤ nused, 0 + nused ≤ # shl(zbrar, n, nused + 1) ⊦ (∃ n0. n0 < nused ∧ shl(zbrar, n, nused + 1)[0 + n0].key = key) ↔ key ≠ zbrar[n].key ∧ key ∈ keys(zbrar, 0, nused + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n < # zbrar, nused < # zbrar, sorted(zbrar, 0, nused + 1), n ≤ nused ⊦ (∃ n0. n0 < nused ∧ (n0 < n ⊃ zbrar[n0];zbrar[n0 + 1]).key = key) ↔ key ≠ zbrar[n].key ∧ key ∈ keys(zbrar, 0, nused + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%n ≤ m ⊦ (m - n) + m0 ≤ n0 ↔ m + m0 ≤ n0 + n; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)%m0 < # zbrar, n ≤ m ⊦ shl(zbrar, n, m)[m0] = (m0 < n ∨ n + (m - n - 1) ≤ m0 ⊃ zbrar[m0];zbrar[m0 + 1]); (s)% ⊦ ¬ m + n < m; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s)% ⊦ m < n → n ≠ 0; (s)%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%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → m ≤ n; (s)% ⊦ # shl(zbrar, n, m) = # zbrar; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)