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Proof step 19 from proof dups-def (previous step)

x0[m0] = (a ' + x0)[n], m0 < # x0, n ≤ # x0 ⊦ m0 + 1 = n, x0 = [], dups(x0), a ∈ x0, n - 1 < # x0 ∧ m0 ≠ n - 1 ∧ x0[m0] = x0[n - 1], ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n]

User applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m1] flags:

Additional information

Used simplifier rules: get-zero-01: ⊦ (a ' + x)[0] = a;%get-nodups: ⊦ m < # x ∧ n < # x ∧ ¬ dups(x) → (x[m] = x[n] ↔ m = n);%get-in: ⊦ m < # x → (¬ x[m] ∈ x ↔ false);%get-succ: ⊦ (a ' + x)[n + 1] = x[n];%get-nodups: ⊦ m < # x ∧ n < # x ∧ ¬ dups(x) → (x[m] = x[n] ↔ m = n);%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%lez: ⊦ 0 ≤ m; from specification nat%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m1