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Proof step 21 from proof ocs-sw (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ x[m] ≠ a ∧ a ≠ b ∧ m < # x → #oc(a, x[m, b]) = #oc(a, x)
Proof step 23:
n < # x, m < # x ⊦ #oc(a, x[m, x[n]]) = #oc(a, x), x[m, x[n]][n] = a, a = x[m], x = [], x[m] ≠ a ∧ a ≠ x[n] ∧ 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[n], m, x, a] flags: <all paths>

Additional information

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