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Proof step 16 from proof ocs-put-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
m ≤ # x0, ∀ m0, b0. x0[m0] ≠ b0 ∧ m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ⊦ m = 0, x0[m - 1] = b, x0[m - 1] = a0, #oc(x0[m - 1], (a0 ' + x0)[m, b]) = #oc(x0[m - 1], x0) - 1

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list