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Proof step 5 from proof remove-refsize0-eq-nonzero (previous step)

st[n] = mkblock(x, lax, 0), n ∈ st, ∀ n. (n ∈ remove-refsize0(st) ↔ n ∈ st) ∧ remove-refsize0(st)[n] = st[n] ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n abort: F discard: T computed: T

Additional information

Used simplifier rules: in-remove-refsize0: ⊦ n ∈ remove-refsize0(st) ↔ n ∈ st ∧ st[n].ref-size ≠ 0;%at-remove-refsize0: ⊦ n ∈ st ∧ st[n].ref-size ≠ 0 → remove-refsize0(st)[n] = st[n];% ⊦ ¬ n ∈ st → st[n] = (∅)[n]; (s)% ⊦ mkblock(x, lax, m).ref-size = m; (s) Used terms: n