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Proof step 15 from proof subsize (previous step)

# y0 + # z < # x0, #oc(a, x0) ≤ #oc(a, y0) + #oc(a, z), ∀ a0. #oc(a0, x0) ≤ #oc(a0, y0) + #oc(a0, z) ⊦ x0 = [], ∃ y. # y < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y))

User applied rule exists right

And obtained the following new goals

Proof step 16:
# y0 + # z < # x0, #oc(a, x0) ≤ #oc(a, y0) + #oc(a, z), ∀ a0. #oc(a0, x0) ≤ #oc(a0, y0) + #oc(a0, z) ⊦ x0 = [], ∃ y. # y < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y)), #(y0 + z) < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y0 + z))

Rule argument

pos: right 2 terms: y0 + z abort: F discard: F computed: F

Additional information

Used terms: y0 + z