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Proof step 13 from proof dups-def (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 14:
(a ' + x0)[m] = (a ' + x0)[n], m ≤ # x0, n ≤ # x0 ⊦ m = n, dups(x0), a ∈ x0, ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n], m -1 < # x0 ∧ n -1 < # x0 ∧ m -1 ≠ n -1 ∧ x0[m -1] = x0[n -1]

Rule argument

pos: right 4 terms: m -1, n -1 abort: F discard: F computed: F

Additional information

Used terms: m -1, n -1