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Proof step 5 from proof ocs-sw (previous step)

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]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
m < # x ⊦ m ≠ n → x[m, a][n] = x[n]
Proof step 7:
m < # x ∧ m ≠ n, n < # x, m < # x ⊦ #oc(a, x[m, x[n]][n, x[m]]) = #oc(a, x), x = [], x[n] ≠ a ∧ a ≠ x[m]

Rule argument

spec: list-del inst: name: put-at seq: m < # x ⊦ m ≠ n → x[m, a][n] = x[n] subst: [n, a, m, x] → [n, x[n], m, x] flags: rewrite

Additional information

Used simplifier rules: lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%putset-same: ⊦ n < # x → x[n, x[n]] = x; from specification list-del Used terms: n, x[n], m, x