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Proof step 8 from proof key-range-replace-branches (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 9:
# zbrar0 = # zbrar1, n < m, n < # zbrar1, m ≤ # zbrar1, ∀ n0. n0 ≤ n ∧ n0 ≤ m → zbrar0[n0].key = zbrar1[n0].key ⊦ (n + 1 = m ⊃ ri |> zbrar0[n - 1].key;(ri |> zbrar0[n - 1].key) |≤ zbrar0[n].key) = (n = 0 ⊃ (m = 1 ⊃ ri;ri |≤ zbrar1[0].key);(n + 1 = m ⊃ ri |> zbrar1[n - 1].key;(ri |> zbrar1[n - 1].key) |≤ zbrar1[n].key)), n = 0

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ 0 ≤ m; (s)