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Proof step 10 from proof refsize-cons-in-ri-0 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 = (range ri | m).size, gstore[m] = groupblock(gndx, lax, 0), @(m, m0, m1) ∈ ri, m ∈ gstore, ri.lebs ⊆ dom gstore ⊦ m1 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ groupblock(gndx, lax, m).ref-size = m; (s)