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Proof step 9 from proof inv-read-gblock (previous step)

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), maxino2 ∈ gstore2, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪if err0 = ESUCCESS then ax := ax + gstore2[maxino2].ladrlist.to-adr(maxino2), gndx := gndx + gstore2[maxino2].gndlist else ax := ?, gndx := ?⟫ true, dups(nx), maxino2 ∈ ginval2

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 10:
err0 = ESUCCESS, 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), maxino2 ∈ gstore2, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪ax := ax + gstore2[maxino2].ladrlist.to-adr(maxino2), gndx := gndx + gstore2[maxino2].gndlist⟫ true, dups(nx), maxino2 ∈ ginval2
Proof step 11:
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), maxino2 ∈ gstore2, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪ax := ?, gndx := ?⟫ true, err0 = ESUCCESS, dups(nx), maxino2 ∈ ginval2

Rule argument

pos: right 1

Proof step 9 from proof inv-read-gblock (previous step)

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

Additional information