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Proof step 2 from proof key-range-replace-branches (previous step)

# zbrar0 = # zbrar1, n < m, m ≤ # zbrar0, ∀ n0. n0 ≤ n ∧ n0 ≤ m → zbrar0[n0].key = zbrar1[n0].key ⊦ key-range(zbrar0, n, m, ri) = key-range(zbrar1, n, m, ri)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# zbrar0 = # zbrar1, n < m, n < # zbrar1, m ≤ # zbrar1, ∀ n0. n0 ≤ n ∧ n0 ≤ m → zbrar0[n0].key = zbrar1[n0].key ⊦ key-range(zbrar0, n, m, ri) = key-range(zbrar1, n, m, ri)

Rule argument

Additional information

Used simplifier rules: lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic