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Proof step 17 from proof sub-del-any (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: ocs-del-length seq: ⊦ a ≠ b → #oc(a, x -l b) = #oc(a, x) subst: [b, x, a] → [a, x, a0] flags: <all paths>

Additional information

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