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Proof step 14 from proof refine-allocate-gnd (previous step)

ginval = ginval3, gstore = gstore3, nx = nx0, pstate = readonly, remove-refsize0(gstore2) = gstore2, remove-refsize0(istore1) = istore1, elems nx ∩ ginval = ∅, elems nx0 ∩ ginval3 = ∅, dom gstore2 ∪ elems nx = dom gstore ∪ ginval, dom gstore2 ∪ elems nx0 = dom gstore3 ∪ ginval3, btree-inv(rp, adr, istore2, istore1, zh, fi, fi0), inv(gstore \ ginval), inv(gstore3 \ ginval3), inv(gstore2), inv(istore2), inv(istore1), ri-inj(fi), ri-inj(fi0), istore1 ⊑ istore2, pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore, pstate.journal? → nx0 ≠ [] ∧ pstate.lnum = nx0.last ∧ pstate.lnum ∈ gstore3, pstate.index? → pstate.lnum ∈ istore2 ⊦ ⟨index_allocate_gnd#(fi, fi0, gstore2, maxino0, ginval2, true; ginval3, gstore3, nx0; err0)⟩ (ginval3 = ginval ∧ gstore3 = gstore ∧ nx0 = nx ∧ EROFS = err0), dups(nx0), dups(nx)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
ginval = ginval3, gstore = gstore3, nx = nx0, remove-refsize0(istore1) = istore1, remove-refsize0(gstore2) = gstore2, elems nx0 ∩ ginval3 = ∅, dom gstore2 ∪ elems nx0 = dom gstore3 ∪ ginval3, btree-inv(rp, adr, istore2, istore1, zh, fi, fi0), inv(istore1), inv(istore2), inv(gstore2), inv(gstore3 \ ginval3), ri-inj(fi0), ri-inj(fi), istore1 ⊑ istore2, ¬ dups(nx0) ⊦ ⟨index_allocate_gnd#(fi, fi0, gstore2, maxino0, ginval2, true; ginval3, gstore3, nx0; err0)⟩ (ginval3 = ginval ∧ gstore3 = gstore ∧ nx0 = nx ∧ EROFS = err0)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ readonly.journal?; (s)% ⊦ ¬ readonly.index?; (s)