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Proof step 10 from proof subsize (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 11:
a ∈ y, #oc(a, x0) < #oc(a, y), # y ≤ # x0, ∀ a0. #oc(a0, a ') + #oc(a0, x0) ≤ #oc(a0, y) ⊦ x0 = [], ∃ y. # y < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y))

Rule argument

fma: a ∈ y

Additional information

Used simplifier rules: notin: ⊦ ¬ a ∈ x → #oc(a, x) = 0;%less-1: ⊦ ¬ n < 0; from specification nat-basic1 Used formulas: a ∈ y