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Proof step 7 from proof subdel (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a0 abort: F discard: F computed: F

Additional information

Used simplifier rules: oc-del: ⊦ ¬ #oc(a, x) < #oc(a, x -1l b);%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat Used terms: a0