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Proof step 15 from proof size-0 (previous step)

(range ri | m).size = 0, ri[key] = @(m, m0, m1), inv(gstore \ ginval), key-cons(key, nodes(gstore ↓, ginval)[@(m, m0, m1)]), key ∈ ri, m ∈ gstore, @(m0, m1) ∈ gstore[m] ↓.ladrlist ⊦ m ∈ ginval

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
(range ri | m).size = 0, ri[key] = @(m, m0, m1), gstore[m] ↓ = groupblock(gndx, lax, m2), inv(gstore \ ginval), key-cons(key, nodes(gstore ↓, ginval)[@(m, m0, m1)]), @(m0, m1) ∈ lax, m ∈ gstore, key ∈ ri ⊦ m ∈ ginval

Rule argument

spec: ablock, inst: group-store name: elim seq: ⊦ (gndx = gblock.gndlist ∧ lax = gblock.ladrlist) ∧ m = gblock.ref-size ↔ gblock = groupblock(gndx, lax, m) subst: [gblock, gndx, lax, m] → [gstore[m] ↓, gndx, lax, m2]

Additional information

Used simplifier rules: ⊦ (gndx = gblock.gndlist ∧ lax = gblock.ladrlist) ∧ m = gblock.ref-size ↔ gblock = groupblock(gndx, lax, m); (elim) Used terms: gstore[m] ↓, gndx, lax, m2