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Proof step 2 from proof rev-at (previous step)

n < # ar ⊦ rev(ar)[n] = ar[# ar - n + 1], # ar = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
# ar = n0 + n + 1 ∧ n + 1 ≤ # ar, n < # ar ⊦ rev(ar)[n] = ar[n0], # ar = 0

Rule argument

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