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Proof step 12 from proof inv-store (previous step)

syn0 ↔ syn2, syn ↔ syn1, gstore2[adr2] = mkgnode(nd, m, boolvar, boolvar0), adr0 = adr2, adr3 = adr1, ginval = ginval1, ginval0 = ginval2, gstore = gstore1, gstore0 = gstore2, remove-refsize0(gstore1) = gstore1, elems nx ∩ ginval2 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval2, inv(gstore1), inv(gstore2 \ ginval2), key-cons(key, nd), ri-inj(fi0), ri-inj(fi), adr2 ∈ gstore2 ⊦ ⟪fi[key] := adr0⟫ ri-inj(fi), dups(nx), key ∈ fi, adr2 ∈ fi

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ri-inj(ri) ∧ ¬ adr ∈ ri → ri-inj(ri[key, adr]); (ss)