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Proof step 5 from proof ocs-put-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
(a0 ' + x0)[m] = a0, m ≤ # x0, ∀ m0, b0. m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ∨ x0[m0] = b0 ⊦ #oc(a0, (a0 ' + x0)[m, b]) = #oc(a0, x0), a0 = b

Rule argument

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)