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Proof step 9 from proof refsize-cons-in-ri-0 (previous step)

gstore[m] = groupblock(gndx, lax, 0), m ∈ gstore, @(m, m0, m1) ∈ ri, ri.lebs ⊆ dom gstore, ∀ n. n ∈ gstore → gstore[n].ref-size = (range ri | n).size ⊦ m1 = 0

User applied rule all left

And obtained the following new goals

Proof step 10:
gstore[m] = groupblock(gndx, lax, 0), @(m, m0, m1) ∈ ri, m ∈ gstore, ri.lebs ⊆ dom gstore, m ∈ gstore → gstore[m].ref-size = (range ri | m).size ⊦ m1 = 0

Rule argument

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

Additional information

Used terms: m