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Proof step 11 from proof ocs-put-01 (previous step)

m0 < # x0, ∀ m, a, b0. m < # x0 → x0[m] = a ∨ a = b0 ∨ #oc(a, x0[m, b0]) = #oc(a, x0) ⊦ #oc(a0, x0[m0, b]) = #oc(a0, x0), a0 = b, x0[m0] = a0, x0 = []

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: m0, a0, b

Additional information

Used terms: m0, a0, b