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Proof step 11 from proof inv-remove (previous step)

jvalid ↔ key ∈ fi, gstore = gstore1, gstore0 = gstore2, remove-refsize0(gstore1) = gstore1, elems nx ∩ jro = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ jro, fi0.inos < pmaxino, fi.inos < jmaxino, p_os < pmaxino, inv(gstore1), inv(gstore2 \ jro), refsize-cons(gstore1, range fi0), refsize-cons(gstore2, range fi), ri-inj(fi0), ri-inj(fi), pmaxino ≤ jmaxino, fi0.lebs ⊆ dom gstore1, fi.lebs ⊆ dom gstore2, jvalid1 ∧ nx ≠ [] → valid-journal-head(nx, gstore2, jro) ⊦ [if jvalid then {adr := fi[key]; fi := fi -- key} else adr := ?] ( ri-inj(fi) → ⟪if jvalid then let n = ? in {index_get_gblock_refsize#(adr.lnum, fi, fi0, jro, gstore2, gstore1, nx, pmaxino, p_os, jvalid0; ; n); index_set_gblock_refsize#(adr.lnum, n - adr.size, fi, fi0, jro, gstore1, nx, pmaxino, p_os, jvalid0; gstore2)}⟫ ( fi.inos < jmaxino ∧ (jvalid1 ∧ nx ≠ [] → valid-journal-head(nx, gstore2, jro)) ∧ fi.lebs ⊆ dom gstore2 ∧ refsize-cons(gstore2, range fi) ∧ inv(gstore2 \ jro) ∧ dom gstore1 ∪ elems nx = dom gstore2 ∪ jro)), pmaxino = 0, dups(nx)

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

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