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Proof step 11 from proof ocs-put (previous step)

m = m0 + 1, m ≤ # x0, ∀ m, a. m < # x0 → #oc(a, x0[m, a]) = #oc(a, x0) + 1 ∨ x0[m] = a ∨ x0 = [] ⊦ #oc(a0, (a0 ' + x0)[m, a0]) = 2 + #oc(a0, x0), x0[m0] = a0, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m0 < # x0, ∀ m, a. m < # x0 → #oc(a, x0[m, a]) = #oc(a, x0) + 1 ∨ x0[m] = a ⊦ #oc(a0, x0[m0, a0]) = #oc(a0, x0) + 1, x0[m0] = a0, x0 = []

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%one: ⊦ #oc(a, a ') = 1;% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)% ⊦ (b ' + x)[n + 1, a] = b + x[n, a]; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)