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Proof step 17 from proof ocs-put-02 (previous step)

m ≤ # x0, ∀ m0, b0. m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ∨ x0[m0] = b0 ⊦ #oc((a0 ' + x0).tail[m - 1], (a0 ' + x0)[m, b]) = #oc((a0 ' + x0).tail[m - 1], x0) - 1, a0 ' + x0 = [], (a0 ' + x0).tail[m - 1] = b, (a0 ' + x0).tail[m - 1] = a0, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
m ≤ # x0, ∀ m0, b0. m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ∨ x0[m0] = b0 ⊦ #oc(x0[m - 1], (a0 ' + x0)[m, b]) = #oc(x0[m - 1], x0) - 1, x0[m - 1] = a0, x0[m - 1] = 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)