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Proof step 7 from proof ocs-sw (
previous step
)
m < # x ∧ m ≠ n, n < # x, m < # x ⊦ #oc(a, x[m, x[n]][n, x[m]]) = #oc(a, x), x = [], x[n] ≠ a ∧ a ≠ x[m]
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 8
:
n < # x, m < # x ⊦ #oc(a, x[m, x[n]][n, x[m]]) = #oc(a, x), x = [], m = n, x[n] ≠ a ∧ a ≠ x[m]
Rule argument
Additional information