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Proof step 10 from proof perm-all (previous step)

∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y) ⊦ #oc(a, x) ≤ #oc(a, y) ∧ #oc(a0, y) ≤ #oc(a0, x)

User applied rule all left

And obtained the following new goals

Proof step 11:
a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y), ∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y) ⊦ #oc(a, x) ≤ #oc(a, y) ∧ #oc(a0, y) ≤ #oc(a0, x)

Rule argument

pos: left 1 terms: a abort: F discard: F computed: F

Additional information

Used terms: a