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Proof step 10 from proof ocs-put-02 (previous step)

(a0 ' + x0)[m] = a0, m = m1 + 1, m ≤ # x0, ∀ m0, b0. m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ∨ x0[m0] = b0 ⊦ 1 + #oc(a0, x0[m1, b]) = #oc(a0, x0), a0 = b, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
m1 < # x0, ∀ m0, b0. m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ∨ x0[m0] = b0 ⊦ 1 + #oc(x0[m1], x0[m1, b]) = #oc(x0[m1], x0), x0[m1] = b, x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ (a ' + x)[n + 1] = x[n]; (s)