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Proof step 6 from proof rev-at-h (previous step)

# 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]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
# ar = n2 + n1 + 1 ∧ n1 + 1 ≤ # ar, # 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[n2]][m] = ar[n0]

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# ar, n1 + 1, n] flags:

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # ar, n1 + 1, n