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Proof step 7 from proof sub-del-any (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ a ≠ b → #oc(a, x -l b) = #oc(a, x)
Proof step 9:
¬ a0 ≠ a, ∀ a. #oc(a, x) ≤ #oc(a, y) ⊦ #oc(a0, x -l a) ≤ #oc(a0, y -l a) ∧ #oc(a, x) ≤ #oc(a, y)
Proof step 11:
a0 ≠ a, ∀ a. #oc(a, x) ≤ #oc(a, y) ⊦ #oc(a0, x) ≤ #oc(a0, y -l a) ∧ #oc(a, x) ≤ #oc(a, y)

Rule argument

spec: inst: name: ocs-del-length seq: ⊦ a ≠ b → #oc(a, x -l b) = #oc(a, x) subst: [b, x, a] → [a, x, a0] flags: rewrite, current

Additional information

Used terms: a, x, a0