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Proof step 3 from proof ocs-put-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m ≤ # x0, ∀ m, a, b. m < # x0 → x0[m] = a ∨ a = b ∨ #oc(a, x0[m, b]) = #oc(a, x0) ∨ x0 = [] ⊦ (a0 ' + x0)[m] = a, a = b, #oc(a, (a0 ' + x0)[m, b]) = #oc(a, a0 ') + #oc(a, x0)

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list