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Proof step 5 from proof adr-in-ri-leb-notin-ginval (previous step)

ri[key] = @(m0, m1, m2), key ∈ ri, m0 ∈ ginval, ∀ key. key ∈ ri → ri[key] ∈ nodes(gstore ↓, ginval) ∧ key-cons(key, nodes(gstore ↓, ginval)[ri[key]]) ⊦ # ginval = 0, ginval = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
ri[key] = @(m0, m1, m2), m0 ∈ ginval, key ∈ ri, ∀ key0. key0 ∈ ri → ri[key0] ∈ gstore ↓ ∧ ¬ ri[key0].lnum ∈ ginval ∧ key-cons(key0, nodes(gstore ↓, ginval)[ri[key0]]) ⊦ # ginval = 0, ginval = ∅

Rule argument

Additional information

Used simplifier rules: nodes-gstore-in: ⊦ adr ∈ nodes(gstore, ginval) ↔ ¬ ¬(adr ∈ gstore ∧ ¬ adr.lnum ∈ ginval);