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Proof step 6 from proof ocs-dups-ex (previous step)

⊦ ∃ a0. (a0 ∈ x0 ∨ a0 = a ∨ a0 ∈ y ∨ a0 ∈ z) ∧ #oc(a0, x0) + #oc(a0, a ') + #oc(a0, y) + #oc(a0, a ') + #oc(a0, z) ≠ 1

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: a abort: F discard: T computed: T

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-num-03: n < n0 ⊦ (* n) ≠ m0 + (* n0); from specification nat Used terms: a