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Proof step 8 from proof subdel-01 (previous step)

#oc(a0, (y + y0) -1l a) < #oc(a0, x), a ∈ y0, a0 ∈ x, ∀ a1. #oc(a1, x) ≤ #oc(a1, y) + #oc(a1, y0 -1l a) ⊦

User applied rule all left

And obtained the following new goals

Proof step 9:
#oc(a0, (y + y0) -1l a) < #oc(a0, x), a ∈ y0, a0 ∈ x, #oc(a0, x) ≤ #oc(a0, y) + #oc(a0, y0 -1l a), ∀ a1. #oc(a1, x) ≤ #oc(a1, y) + #oc(a1, y0 -1l a) ⊦

Rule argument

pos: left 4 terms: a0 abort: F discard: F computed: F

Additional information

Used terms: a0