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Proof step 20 from proof key-range-replace-branches (previous step)

# 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

User applied rule all left

And closed the goal!

Rule argument

pos: left 10 terms: m0 + 1 abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ ¬ m < n → ¬ m + m0 < n; (s)% ⊦ m < n → m ≤ n; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat Used terms: m0 + 1