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Proof step 5 from proof keys-shift-left (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n < # zbrar, nused < # zbrar, sorted(zbrar, 0, nused + 1), n ≤ nused ⊦ key ∈ keys(shl(zbrar, n, nused + 1), 0, nused) ↔ key ≠ zbrar[n].key ∧ key ∈ keys(zbrar, 0, nused + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ key ∈ ks -- key0 ↔ key ≠ key0 ∧ key ∈ ks; (s)