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Proof step 5 from proof inv-remove (previous step)

syn ↔ syn1, syn0 ↔ syn2, 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), ri-inj(fi), ri-inj(fi0) ⊦ ⟪if key ∈ fi then {adr := fi[key]; fi := fi -- key} else adr := ?⟫ ri-inj(fi), dups(nx)

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 6:
syn0 ↔ syn2, syn ↔ syn1, 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), ri-inj(fi0), ri-inj(fi), key ∈ fi ⊦ ⟪adr := fi[key]; fi := fi -- key⟫ ri-inj(fi), dups(nx)
Proof step 8:
syn0 ↔ syn2, syn ↔ syn1, 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), ri-inj(fi0), ri-inj(fi) ⊦ ⟪adr := ?⟫ ri-inj(fi), dups(nx), key ∈ fi

Proof step 5 from proof inv-remove (previous step)

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

Additional information