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Proof step 18 from proof subrec (previous step)

a ∈ y, #oc(a, y) ≤ #oc(a, x), ∀ a0. #oc(a0, x) ≤ #oc(a0, y -1l a) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a

Additional information

Used simplifier rules: del1: ⊦ a ∈ x → #oc(a, x -1l a) = #oc(a, x) -1;%in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0;%pred-03: ⊦ n -1 = n - 1; from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%ls-02: n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; from specification nat%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%le-ls-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 Used terms: a