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Proof step 18 from proof ocs-sw (previous step)

m < # x, n < # x ⊦ #oc(x[m], x[m, x[n]]) + 1 = #oc(x[m], x), x[m, x[n]][n] = x[m], x[n] = x[m], x = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ x[m] = a ∧ a ≠ b ∧ m < # x → #oc(a, x[m, b]) = #oc(a, x) -1

Rule argument

spec: <toplevel> name: ocs-put-02 seq: ⊦ x[m] = a ∧ a ≠ b ∧ m < # x → #oc(a, x[m, b]) = #oc(a, x) -1 subst: [b, m, x, a] → [x[n], m, x, x[m]] flags: <all paths>

Additional information

Used simplifier rules: oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;% ⊦ n -1 = n - 1; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m < # x → (x[m] ∈ x ↔ true); (ss)%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic Used terms: x[n], m, x, x[m]