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Proof step 3 from proof lastn-length-01 (previous step)

n ≤ # x ⊦ # x - (# x - n) = n

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)%n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0; (s)% ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ n < n; (s)%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ ¬ m + n < m; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x, # x - n, n0