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Proof step 6 from proof at-range (previous step)

n < nused, n < # zbrar, key ∈ key-range(zbrar, n, nused, ri), nused ≤ # zbrar ⊦ key-range(zbrar, n, nused, ri)[key] = ri[key], n ≠ 0 ∧ n + 1 < nused

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: n ≠ 0

Additional information

Used simplifier rules: range-last: ⊦ n ≠ 0 ∧ n < # zbrar → key-range(zbrar, n, n + 1, ri) = ri |> zbrar[n - 1].key;% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)% ⊦ key0 ∈ ri |> key1 ↔ key0 ∈ ri ∧ key1 < key0; (s)% ⊦ key1 < key0 ∧ key0 ∈ ri → (ri |> key1)[key0] = ri[key0]; (s) Used formulas: n ≠ 0