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Proof step 10 from proof put-del1-in (previous step)

m0 < # x0, ∀ m, a. x0[m] ≠ a ∧ m < # x0 → (a ∈ x0[m, a] -1l a ↔ a ∈ x0) ⊦ x0 = [], a = a0, x0[m0] = a, a ∈ x0[m0, a] -1l a ↔ a ∈ x0

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: m0, a abort: F discard: F computed: T

Additional information

Used terms: m0, a