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Proof step 10 from proof del-02 (previous step)

a ∈ y, a ∈ x, ∀ a0. #oc(a0, x -1l a) ≤ #oc(a0, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x -1l a) ⊦ #oc(a1, y) ≤ #oc(a1, x)

User applied rule all left

And obtained the following new goals

Proof step 11:
a ∈ x, a ∈ y, #oc(a1, y -1l a) ≤ #oc(a1, x -1l a), ∀ a0. #oc(a0, x -1l a) ≤ #oc(a0, y -1l a) ⊦ #oc(a1, y) ≤ #oc(a1, x)

Rule argument

pos: left 4 terms: a1 abort: F discard: T computed: F

Additional information

Used terms: a1