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Proof step 22 from proof range-put (previous step)

nused0 < # zbrar ⊦ nused0 = 0, zbrar[0].key < key, (key0 ∈ ri[key, adr] |≤ zbrar[0].key ↔ key0 = key ∨ key0 ∈ key-range(zbrar, 0, nused0 + 1, ri)) ∧ (key0 ∈ ri[key, adr] |≤ zbrar[0].key → ri[key, adr][key0] = key-range(zbrar, 0, nused0 + 1, ri)[key, adr][key0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
nused0 < # zbrar ⊦ nused0 = 0, zbrar[0].key < key, (key0 = key ∨ key0 ∈ ri ∧ ¬ zbrar[0].key < key0 → key0 ∈ (ri |≤ zbrar[0].key)[key, adr]) ∧ (key0 ∈ (ri |≤ zbrar[0].key)[key, adr] → ri[key, adr][key0] = (ri |≤ zbrar[0].key)[key, adr][key0])

Rule argument

Additional information

Used simplifier rules: range-first: ⊦ nused ≠ 0 ∧ nused < # zbrar → key-range(zbrar, 0, nused + 1, ri) = ri |≤ zbrar[0].key;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ key0 ∈ ri |≤ key1 ↔ key0 ∈ ri ∧ ¬ key1 < key0; (s)% ⊦ key ≠ key0 → (key ∈ ri[key0, adr] ↔ key ∈ ri); (s)% ⊦ ¬ key1 < key0 → ri[key0, adr] |≤ key1 = (ri |≤ key1)[key0, adr]; (s)