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Proof step 17 from proof ocs-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
m0 < # x0, ∀ m, a1. m < # x0 → x0[m] = a1 ∨ #oc(a1, x0[m, a1]) = #oc(a1, x0) + 1 ⊦ #oc(a, x0[m0, a]) = 1 + #oc(a, x0), x0[m0] = a, a = a0, x0 = []

Rule argument

Additional information

Used simplifier rules: oc-n: ⊦ a ≠ b → #oc(a, b ' + x) = #oc(a, x);%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ a + x = a ' + x; from specification list%put-succ: ⊦ (b ' + x)[n + 1, a] = b + x[n, a]; from specification list-del