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Proof step 9 from proof rev-at (previous step)

# ar = n0 + n + 1, n < # ar ⊦ # ar = 0, ar[# ar - n + 1] = ar[n0]

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic2%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # ar, n + 1, n1