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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
n0 + m = n1 + n2, # ar = n0 + m + 1, # ar = n2 + n1 + 1, m < n2 + n1, m < n1, m < # ar, n1 < # ar, ∀ ar0, n, m0. # ar0 = n + m0 + 1 ∧ m0 < n1 ∧ n1 ≤ # ar0 ∧ m0 < # ar0 → # ar0 = 0 ∨ rev(ar0, n1)[m0] = ar0[n] ⊦ n2 + n1 = 0, n0 = 0, n1 = 0, m = n1, rev(ar, n1)[m] = ar[n0], # ar = 0

Rule argument

Additional information

Used simplifier rules: l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%lels-03: ⊦ m < n → m ≤ n; from specification nat%l-06: ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; from specification nat-basic2%one-03: ⊦ 1 = m + 1 ↔ m = 0; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%succ: ⊦ m +1 = m + 1; from specification nat-basic2%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat