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Proof step 9 from proof ocs-put-01 (previous step)

m ≤ # x0, ∀ m, a, b. m < # x0 → #oc(a, x0[m, b]) = #oc(a, x0) ∨ x0[m] = a ∨ a = b ∨ x0 = [] ⊦ #oc(a0, (a0 ' + x0)[m, b]) = 1 + #oc(a0, x0), a0 ' + x0 = [], (a0 ' + x0).tail[m - 1] = a0, a0 = b, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
m ≤ # x0, ∀ m, a, b. m < # x0 → #oc(a, x0[m, b]) = #oc(a, x0) ∨ x0[m] = a ∨ a = b ∨ x0 = [] ⊦ #oc(a0, (a0 ' + x0)[m, b]) = 1 + #oc(a0, x0), x0[m - 1] = a0, a0 = b, m = 0

Rule argument

Additional information

Used simplifier rules: ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)