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Proof step 15 from proof key-range-replace-branches (previous step)

# zbrar0 = # zbrar1, n < m, n < # zbrar1, m ≤ # zbrar1, ∀ n0. n0 ≤ n ∧ n0 ≤ m → zbrar0[n0].key = zbrar1[n0].key ⊦ n + 1 = m, n = 0, (ri |> zbrar0[n - 1].key) |≤ zbrar0[n].key = (n + 1 = m ⊃ ri |> zbrar1[n - 1].key;(ri |> zbrar1[n - 1].key) |≤ zbrar1[n].key)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# zbrar0 = # zbrar1, n + 1 < m, n < m, n < # zbrar1, n + 1 < # zbrar1, m ≤ # zbrar1, ∀ n0. n0 ≤ n ∧ n0 ≤ m → zbrar0[n0].key = zbrar1[n0].key ⊦ n + 1 = m, n = 0, (ri |> zbrar0[n - 1].key) |≤ zbrar0[n].key = (ri |> zbrar1[n - 1].key) |≤ zbrar1[n].key

Rule argument

Additional information

Used simplifier rules: lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic