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Proof step 8 from proof remove-refsize0-refsize-cons (previous step)

inv(st), @(m, m0, m1) ∈ as, @(m, m0, m1) ∈ st, as ⊆ dom nodes(st), ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅, @(m, m0, m1) ∈ remove-refsize0(st), dom nodes(st) ⊆ dom nodes(remove-refsize0(st))

User applied rule all left

And obtained the following new goals

Proof step 9:
inv(st), @(m, m0, m1) ∈ st, @(m, m0, m1) ∈ as, as ⊆ dom nodes(st), m ∈ st → st[m].ref-size = (as | m).size ⊦ # as = 0, as = ∅, @(m, m0, m1) ∈ remove-refsize0(st), dom nodes(st) ⊆ dom nodes(remove-refsize0(st))

Rule argument

pos: left 5 terms: m abort: F discard: T computed: F

Additional information

Used terms: m