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Proof step 6 from proof inv-store (previous step)

syn0 ↔ syn2, syn ↔ syn1, gstore2[adr2] = mkgnode(nd, m, boolvar, boolvar0), adr = adr1, adr0 = adr2, 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 ⊦ ⟪if ¬ adr0 ∈ fi ∧ adr0 ∈ gstore0 ∧ key-cons(key, gstore0[adr0].nd) then {syn := key ∈ fi; if syn then adr := fi[key] else adr := ?; fi[key] := adr0} else abort⟫ ri-inj(fi), dups(nx), adr2 ∈ fi

Heuristic symbolic execution applied rule if positive right

And obtained the following new goals

Proof step 7:
syn ↔ syn1, syn0 ↔ syn2, gstore2[adr2] = mkgnode(nd, m, boolvar, boolvar0), adr = adr1, adr0 = adr2, ginval = ginval1, ginval0 = ginval2, gstore = gstore1, gstore0 = gstore2, remove-refsize0(gstore1) = gstore1, elems nx ∩ ginval2 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval2, inv(gstore2 \ ginval2), inv(gstore1), key-cons(key, nd), ri-inj(fi), ri-inj(fi0), adr2 ∈ gstore2 ⊦ ⟪syn := key ∈ fi; if syn then adr := fi[key] else adr := ?; fi[key] := adr0⟫ ri-inj(fi), dups(nx), adr2 ∈ fi

Rule argument

Additional information

Used simplifier rules: ⊦ mkgnode(nd, m, boolvar, boolvar0).nd = nd; (s)