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Proof step 5 from proof rev-at-h (previous step)

∀ ar, n0, m. # ar = n0 + m + 1 ∧ m < n1 ∧ n1 ≤ # ar ∧ m < # ar → # ar = 0 ∨ rev(ar, n1)[m] = ar[n0] ∨ n1 = 0, # ar = n0 + m + 1, m < n1 +1, n1 +1 ≤ # ar, m < # ar ⊦ # ar = 0, rev(ar, n1 +1)[m] = ar[n0], n1 +1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# ar = n0 + m + 1, n1 ≤ n0 + m, m < # ar, n1 < # ar, m ≤ n1, ∀ ar, n0, m. # ar = n0 + m + 1 ∧ m < n1 ∧ n1 ≤ # ar ∧ m < # ar → # ar = 0 ∨ rev(ar, n1)[m] = ar[n0] ∨ n1 = 0 ⊦ # ar = 0, rev(ar, n1)[n1, ar[# ar - n1 + 1]][m] = ar[n0]

Rule argument

Additional information

Used simplifier rules: rev-rec: ⊦ n < # ar → rev(ar, n + 1) = rev(ar, n)[n, ar[# ar - n + 1]];%lep: ⊦ m ≤ m + n; from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2