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Proof step 7 from proof in-keys (previous step)

m + nused0 < # zbrar, m < # zbrar, (m + 1) + nused0 ≤ # zbrar → (key ∈ keys(zbrar, m + 1, nused0) ↔ (∃ n. n < nused0 ∧ zbrar[(m + 1) + n].key = key)) ⊦ key = zbrar[m].key ∨ key ∈ keys(zbrar, m + 1, nused0) ↔ (∃ n. ¬ nused0 < n ∧ zbrar[m + n].key = key)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
key ∈ keys(zbrar, m + 1, nused0) ↔ (∃ n. n < nused0 ∧ zbrar[(m + 1) + n].key = key), m < # zbrar, m + nused0 < # zbrar ⊦ key = zbrar[m].key ∨ key ∈ keys(zbrar, m + 1, nused0) ↔ (∃ n. ¬ nused0 < n ∧ zbrar[m + n].key = key)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)