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Proof step 5 from proof range-le (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
nused ≤ # zbrar ⊦ # zbrar = 1, # zbrar = 0, key-range(zbrar, 0, nused, ri |≤ key) = key-range(zbrar, 0, nused, ri), nused = 1, nused = 0, key < zbrar[0].key

Rule argument

Additional information

Used simplifier rules: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic