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

remove-refsize0(gstore1) = gstore1, elems nx ∩ ginval2 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval2, inv(gstore1), inv(gstore2 \ ginval2), ri-inj(fi1), ri-inj(fi2) ⊦ ⟪choose err0 with ⌊err0⌋ ∨ err0 = ESUCCESS in err := err0; if err = ESUCCESS then syn := key ∈ fi2 else syn := ?⟫ true, dups(nx)

Heuristic symbolic execution applied rule choose right

And obtained the following new goals

Proof step 7:
remove-refsize0(gstore1) = gstore1, elems nx ∩ ginval2 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval2, inv(gstore2 \ ginval2), inv(gstore1), ri-inj(fi2), ri-inj(fi1), ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪err := err0; if err = ESUCCESS then syn := key ∈ fi2 else syn := ?⟫ true, dups(nx)
Proof step 11:
remove-refsize0(gstore1) = gstore1, elems nx ∩ ginval2 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval2, inv(gstore2 \ ginval2), inv(gstore1), ri-inj(fi2), ri-inj(fi1) ⊦ dups(nx), ∃ err0. ⌊err0⌋ ∨ err0 = ESUCCESS

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

Heuristic symbolic execution applied rule choose right

And obtained the following new goals

Rule argument

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