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Proof step 12 from proof refine-allocate-gnd (previous step)

ginval = ginval3, gstore = gstore3, nx = nx0, remove-refsize0(gstore2) = gstore2, remove-refsize0(istore1) = istore1, elems nx0 ∩ ginval3 = ∅, dom gstore2 ∪ elems nx0 = dom gstore3 ∪ ginval3, btree-inv(rp, adr, istore2, istore1, zh, fi, fi0), inv(gstore3 \ ginval3), inv(gstore2), inv(istore2), inv(istore1), ri-inj(fi), ri-inj(fi0), istore1 ⊑ istore2, ¬ dups(nx0), pstate.journal? → nx0 ≠ [] ∧ pstate.lnum = nx0.last ∧ pstate.lnum ∈ gstore3, pstate.index? → pstate.lnum ∈ istore2 ⊦ [if pstate = readonly then err := EROFS else {fail_or_success#(; ; err); if err = ESUCCESS then choose n with ¬ n ∈ gstore in gstore[n] := groupblock([], [], 0), nx := nx + n, ginval := ginval -- n, pstate := journal(n)}] ( inv(gstore \ ginval) ∧ elems nx ∩ ginval = ∅ ∧ ¬ dups(nx) ∧ dom gstore2 ∪ elems nx = dom gstore ∪ ginval ∧ (pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore) ∧ (pstate.index? → pstate.lnum ∈ istore2) → ⟨index_allocate_gnd#(fi, fi0, gstore2, maxino0, ginval2, true; ginval3, gstore3, nx0; err0)⟩ (ginval3 = ginval ∧ gstore3 = gstore ∧ nx0 = nx ∧ err = err0))

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

pos: right 1

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