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Proof step 16 from proof rev-at-h (previous step)

# ar = n2 + n1 + 1, n1 < # ar, ∀ ar, n0, m. # ar = n0 + m + 1 ∧ m < # ar ∧ m < n1 ∧ n1 ≤ # ar → rev(ar, n1)[m] = ar[n0] ⊦ rev(ar, n1)[n1, ar[n2]][n1] = ar[n2], n1 < # rev(ar, n1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# ar = n2 + n1 + 1, n1 < # ar, # rev(ar, n1) < # ar, # rev(ar, n1) ≤ n1, # rev(ar, n1) ≤ n2 + n1, ∀ ar, n0, m. # ar = n0 + m + 1 ∧ m < # ar ∧ m < n1 ∧ n1 ≤ # ar → rev(ar, n1)[m] = ar[n0] ⊦ rev(ar, n1)[n1, ar[n2]][n1] = ar[n2]

Rule argument

Additional information

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