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Proof step 19 from proof rev-at-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
# ar = n0 + m + 1, # ar = n2 + n1 + 1, n0 + m = n1 + n2, m < n2 + n1, m < n1, n1 < # ar, m < # ar, ∀ ar, n0, m. # ar = n0 + m + 1 ∧ m < # ar ∧ m < n1 ∧ n1 ≤ # ar → rev(ar, n1)[m] = ar[n0] ⊦ rev(ar, n1)[m] = ar[n0], m = n1, n0 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → m ≤ n; (s)% ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s)% ⊦ 1 = m + 1 ↔ m = 0; (s)% ⊦ m + 1 ≠ 0; (s)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)