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Proof step 6 from proof ocs-not-more (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# x0 < #oc(a0, x0), a0 ∈ x0, ∀ a1. ¬ # x0 < #oc(a1, x0) ∨ ¬ a1 ∈ x0 ⊦

Rule argument

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1;%oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)