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Proof step 3 from proof ocs-sw (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ x[m] ≠ a ∧ a ≠ b ∧ m < # x → #oc(a, x[m, b]) = #oc(a, x)
Proof step 5:
n < # x, m < # x ⊦ #oc(a, x[m, x[n]][n, x[m]]) = #oc(a, x), x = [], x[m, x[n]][n] ≠ a ∧ a ≠ x[m] ∧ n < # x[m, x[n]]
Proof step 20:
n < # x, m < # x, n < # x[m, x[n]] ⊦ #oc(a, x[m, x[n]]) = #oc(a, x), x[m, x[n]][n] = a, a = x[m], x = []

Rule argument

spec: <toplevel> name: ocs-put-01 seq: ⊦ x[m] ≠ a ∧ a ≠ b ∧ m < # x → #oc(a, x[m, b]) = #oc(a, x) subst: [b, m, x, a] → [x[m], n, x[m, x[n]], a] flags: <all paths>

Additional information

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