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Proof step 3 from proof ocs-put-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
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

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ n -1 = n - 1; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)