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Proof step 2 from proof in-01 (previous step)

∀ a0. #oc(a0, x) ≤ #oc(a0, a ' + y), ∀ a0. #oc(a0, a ' + y) ≤ #oc(a0, x) ⊦ 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: oc-y: ⊦ #oc(a, a ' + x) = #oc(a, x) + 1;%notin: ⊦ ¬ a ∈ x → #oc(a, x) = 0;%a: ⊦ (x + y) + z = x + y + z; from specification list%le-num-06: n0 < n ⊦ ¬ m + (* n) ≤ (* n0); 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