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Proof step 7 from proof look-subrange (previous step)

n < # zbrar, n < nused, key ∈ (n = 0 ⊃ (nused = 1 ⊃ ri;ri |≤ zbrar[0].key);(n + 1 = nused ⊃ ri |> zbrar[n - 1].key;(ri |> zbrar[n - 1].key) |≤ zbrar[n].key)), nused ≤ # zbrar, (key ∈ (n = 0 ⊃ (nused = 1 ⊃ ri;ri |≤ zbrar[0].key);(n + 1 = nused ⊃ ri |> zbrar[n - 1].key;(ri |> zbrar[n - 1].key) |≤ zbrar[n].key)) ⊃ (bv ↔ key ∈ ri) ∧ adr = ri[key] ;¬ bv ) ⊦ (bv ↔ key ∈ (n = 0 ⊃ (nused = 1 ⊃ ri;ri |≤ zbrar[0].key);(n + 1 = nused ⊃ ri |> zbrar[n - 1].key;(ri |> zbrar[n - 1].key) |≤ zbrar[n].key))) ∧ ( key ∈ (n = 0 ⊃ (nused = 1 ⊃ ri;ri |≤ zbrar[0].key);(n + 1 = nused ⊃ ri |> zbrar[n - 1].key;(ri |> zbrar[n - 1].key) |≤ zbrar[n].key)) → adr = ri[key])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n < # zbrar, n < nused, key ∈ (n = 0 ⊃ (nused = 1 ⊃ ri;ri |≤ zbrar[0].key);(n + 1 = nused ⊃ ri |> zbrar[n - 1].key;(ri |> zbrar[n - 1].key) |≤ zbrar[n].key)), nused ≤ # zbrar ⊦ key ∈ ri

Rule argument

Additional information

Used simplifier rules: ⊦ key0 ∈ ri |> key1 ↔ key0 ∈ ri ∧ key1 < key0; (s)% ⊦ key0 ∈ ri |≤ key1 ↔ key0 ∈ ri ∧ ¬ key1 < key0; (s)