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Proof step 17 from proof inv-commit (previous step)

err0 = ESUCCESS, remove-refsize0(gstore1) = gstore1, dom remove-refsize0(gstore2) ∩ ginval4 = ∅, elems nx ∩ ginval4 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval4, inv(gstore2 \ ginval4), inv(gstore1), ri-inj(fi2), ri-inj(fi1), ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪ginval1 := ?⟫ ( inv(remove-refsize0(gstore2) \ ∅) ∧ inv(remove-refsize0(gstore2)) ∧ elems [] ∩ ∅ = ∅ ∧ ¬ dups([]) ∧ dom remove-refsize0(gstore2) ∪ elems [] = dom remove-refsize0(gstore2) ∪ ∅ ∧ remove-refsize0(remove-refsize0(gstore2)) = remove-refsize0(gstore2) ∧ ri-inj(fi2)), dups(nx)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ⌊ESUCCESS⌋ ; (s)% ⊦ gstore \ ∅ = gstore; (s)%int-a: ⊦ (ns ∩ ns0) ∩ ns1 = ns ∩ ns0 ∩ ns1; from specification set-union[nat-set]%int-c: ⊦ ns ∩ ns0 = ns0 ∩ ns; from specification set-union[nat-set]%⊦ inv(gstore \ ns) ∧ dom remove-refsize0(gstore) ∩ ns = ∅ → inv(remove-refsize0(gstore)); (ss)% ⊦ elems [] = ∅; (s)% ⊦ ns ∩ ∅ = ∅; (s)% ⊦ ¬ dups([]); (s)% ⊦ ns ∪ ∅ = ns; (s)%una: ⊦ (ns ∪ ns0) ∪ ns1 = ns ∪ ns0 ∪ ns1; from specification set-union[nat-set]%unc: ⊦ ns ∪ ns0 = ns0 ∪ ns; from specification set-union[nat-set]% ⊦ remove-refsize0(remove-refsize0(gstore)) = remove-refsize0(gstore); (s)