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Proof step 3 from proof inv-remove (previous step)

remove-refsize0(gstore0) = gstore0, elems nx ∩ ginval = ∅, dom gstore0 ∪ elems nx = dom gstore ∪ ginval, inv(gstore \ ginval), inv(gstore0), ri-inj(fi), ri-inj(fi0), ¬ dups(nx) ⊦ ⟪index_index_remove#(key, fi0, ginval, gstore, gstore0, nx, maxino, ginval0, syn; adr, syn0, fi)⟫ ri-inj(fi)

Heuristic unfold applied rule call right

And obtained the following new goals

Proof step 4:
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) ⊦ ⟪syn := key ∈ fi; if syn then {adr := fi[key]; fi := fi -- key} else adr := ?⟫ ri-inj(fi), dups(nx)

Rule argument

pos: right 1

Additional information

Used simplifier rules: remove: ⊦ ;