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Proof step 12 from proof ocs-sw (previous step)

m < # x, n < # x ⊦ #oc(x[m], x[m, x[n]][n, x[m]]) = #oc(x[m], x), x[n] = x[m], x = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ x[m] ≠ a ∧ m < # x → #oc(a, x[m, a]) = #oc(a, x) + 1
Proof step 14:
n < # x, m < # x ⊦ #oc(x[m], x[m, x[n]][n, x[m]]) = #oc(x[m], x), x[n] = x[m], x = [], x[m, x[n]][n] ≠ x[m] ∧ n < # x[m, x[n]]
Proof step 17:
n < # x, m < # x, n < # x[m, x[n]] ⊦ #oc(x[m], x[m, x[n]]) + 1 = #oc(x[m], x), x[n] = x[m], x[m, x[n]][n] = x[m], x = []

Rule argument

spec: <toplevel> name: ocs-put seq: ⊦ x[m] ≠ a ∧ m < # x → #oc(a, x[m, a]) = #oc(a, x) + 1 subst: [m, x, a] → [n, x[m, x[n]], x[m]] flags: <all paths>

Additional information

Used terms: n, x[m, x[n]], x[m]