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Proof step 19 from proof key-range-replace-branches (previous step)

# zbrar0 = # zbrar1, 2 + m0 < m, m0 + 1 < m, 2 + m0 < # zbrar1, m0 + 1 < # zbrar1, m ≤ # zbrar1, m0 ≤ m0 + 1 ∧ m0 ≤ m → zbrar0[m0].key = zbrar1[m0].key, ∀ n0. n0 ≤ m0 + 1 ∧ n0 ≤ m → zbrar0[n0].key = zbrar1[n0].key ⊦ 2 + m0 = m, (ri |> zbrar0[m0].key) |≤ zbrar0[m0 + 1].key = (ri |> zbrar1[m0].key) |≤ zbrar1[m0 + 1].key

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
# zbrar0 = # zbrar1, zbrar0[m0].key = zbrar1[m0].key, 2 + m0 < m, m0 + 1 < m, 2 + m0 < # zbrar1, m0 + 1 < # zbrar1, m ≤ # zbrar1, m0 ≤ m, m0 ≤ # zbrar1, ∀ n0. n0 ≤ m0 + 1 ∧ n0 ≤ m → zbrar0[n0].key = zbrar1[n0].key ⊦ 2 + m0 = m, (ri |> zbrar1[m0].key) |≤ zbrar0[m0 + 1].key = (ri |> zbrar1[m0].key) |≤ zbrar1[m0 + 1].key

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m + m0 < n → ¬ n + n0 < m; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%transitivity: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification oelem[nat-set]%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)