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Proof step 20 from proof subrec (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a0 abort: F discard: T computed: F

Additional information

Used simplifier rules: del1-01: ⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x);% ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: a0