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Proof step 2 from proof keys-shift-left (previous step)

nused < # zbrar, sorted(zbrar, 0, nused + 1), n ≤ nused ⊦ keys(shl(zbrar, n, nused + 1), 0, nused) = keys(zbrar, 0, nused + 1) -- zbrar[n].key

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < # zbrar, nused < # zbrar, sorted(zbrar, 0, nused + 1), n ≤ nused ⊦ keys(shl(zbrar, n, nused + 1), 0, nused) = keys(zbrar, 0, nused + 1) -- zbrar[n].key

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic