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

a = a0, # x0 + 1 < #oc(a, a0 ') + #oc(a, x0), a = a0 ∨ a ∈ x0, ∀ a1. ¬ # x0 < #oc(a1, x0) ∨ ¬ a1 ∈ x0 ⊦ #oc(a, a0 ') + #oc(a, x0) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# 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-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2