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

err = ESUCCESS, err = err0, ginval = ginval4, gstore = gstore1, gstore0 = crash(gstore2, gstore1, ginval4, nx), remove-refsize0(gstore1) = gstore1, elems nx ∩ ginval4 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval4, inv(gstore1), inv(gstore2 \ ginval4), ri-inj(fi0), ri-inj(fi1), ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪maxino0 := maxino, syn := true, nx0 := nx, ginval1 := ginval0, fi := fi0⟫ ( err = ESUCCESS → syn ∧ inv(gstore0 \ ginval) ∧ inv(gstore) ∧ elems nx ∩ ginval = ∅ ∧ ¬ dups(nx) ∧ dom gstore ∪ elems nx = dom gstore0 ∪ ginval ∧ remove-refsize0(gstore) = gstore ∧ ri-inj(fi) ∧ ri-inj(fi0)), dups(nx)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ⌊ESUCCESS⌋ ; (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]%⊦ inv(gstore \ ginval) ∧ inv(gstore0) ∧ dom gstore0 ∪ elems nx = dom gstore ∪ ginval → inv(crash(gstore, gstore0, ginval, nx) \ ginval); (ss)% ⊦ dom crash(gstore, gstore0, ginval, nx) = dom gstore ∪ ginval; (s)% ⊦ ns ∪ ns0 = ns0 ↔ ns ⊆ ns0; (s)% ⊦ ns ⊆ ns ∪ ns0; (s)