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Proof step 10 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)[n1, ar[n2]][m] = ar[n0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# ar = n2 + n1 + 1, 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 ⊦ rev(ar, n1)[n1, ar[n2]][n1] = ar[n2], # ar = 0

Rule argument

Additional information

Used simplifier rules: l-02: ⊦ ¬ m + n < m; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic2%ler: ⊦ m ≤ m; from specification nat%lep: ⊦ m ≤ m + n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2