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Proof step 9 from proof ocs-not-more (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a

Additional information

Used simplifier rules: ⊦ ¬ m < n → ¬ m + m0 < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: a