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Proof step 11 from proof dups-def (previous step)

n < # x0 ⊦ x0 = [], dups(x0), ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n], ∃ m, n0. ¬ # x0 < m ∧ ¬ # x0 < n0 ∧ m ≠ n0 ∧ (x0[n] ' + x0)[m] = (x0[n] ' + x0)[n0]

User applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: 0, n +1 abort: F discard: F computed: F

Additional information

Used simplifier rules: get-nodups: ⊦ m < # x ∧ n < # x ∧ ¬ dups(x) → (x[m] = x[n] ↔ m = n);%get-zero-01: ⊦ (a ' + x)[0] = a;%get-succ: ⊦ (a ' + x)[n + 1] = x[n];%less-1: ⊦ ¬ n < 0; from specification nat-basic1%succ: ⊦ m +1 = m + 1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: 0, n +1