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Proof step 5 from proof ocs-not-more (previous step)

1 + # x0 < #oc(a, a0 ') + #oc(a, x0), ∀ a1. ¬ # x0 < #oc(a1, x0) ∨ ¬ a1 ∈ x0, a = a0 ∨ a ∈ x0 ⊦

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
a = a0, 1 + # x0 < #oc(a, a0 ') + #oc(a, x0), ∀ a1. ¬ # x0 < #oc(a1, x0) ∨ ¬ a1 ∈ x0, a = a0 ∨ a ∈ x0 ⊦
Proof step 8:
1 + # x0 < #oc(a, a0 ') + #oc(a, x0), ∀ a1. ¬ # x0 < #oc(a1, x0) ∨ ¬ a1 ∈ x0, a = a0 ∨ a ∈ x0 ⊦ a = a0

Rule argument

fma: a = a0

Additional information

Used formulas: a = a0