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Proof step 5 from proof dups-def (previous step)

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

User applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: m + 1, n + 1 abort: F discard: F computed: F

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];%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%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: m + 1, n + 1