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Proof step 22 from proof rmpos-in (
previous step
)
x0[m] ∈ x0 -1l m, m < # x0, x0[m] ∈ x0, ∀ n0, a1. n0 < # x0 → (a1 ∈ x0 -1l n0 ↔ a1 ≠ x0[n0] ∧ a1 ∈ x0) ⊦ x0 = [], a0 ∈ x0, dups(x0)
User applied rule all left
And closed the goal!
Rule argument
pos: left 4 terms: m, x0[m] abort: F discard: F computed: T
Additional information
Used terms: m, x0[m]