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Proof step 14 from proof del (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 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-03: n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat Used terms: a