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Proof step 11 from proof lebs-del-subseteq-set (previous step)

@(m, m0, m1) ∈ as, ∀ n. (∃ adr. adr.lnum = n ∧ adr ∈ as) → n ∈ ns ⊦ # as = 0, @(m, m0, m1) = adr, as = ∅, m ∈ ns, (as -- adr).lebs ⊆ as.lebs

User applied rule all left

And obtained the following new goals

Proof step 12:
@(m, m0, m1) ∈ as, (∃ adr. adr.lnum = m ∧ adr ∈ as) → m ∈ ns ⊦ # as = 0, @(m, m0, m1) = adr, as = ∅, m ∈ ns, (as -- adr).lebs ⊆ as.lebs

Rule argument

pos: left 2 terms: m abort: F discard: T computed: T

Additional information

Used terms: m