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Proof step 10 from proof rev-at-h (previous step)

# ar = n0 + m + 1, # ar = n2 + n1 + 1, n1 < # ar, m < # ar, n1 ≤ n0 + m, m ≤ n1, n1 + 1 ≤ # 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]][m] = ar[n0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# ar = n2 + n1 + 1, # ar = n0 + m + 1, n0 + m = n1 + n2, m < # ar, n1 < # ar, m ≤ n1 + n2, m ≤ n2 + n1, m ≤ 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]][m] = ar[n0]

Rule argument

Additional information

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