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Proof step 41 from proof recovery (previous step)

err2 = err1, remove-refsize0(gstore1) = gstore1, remove-refsize0(istore1) = istore1, elems nx1 ∩ ginval3 = ∅, dom gstore1 ∪ elems nx1 = dom gstore4 ∪ ginval3, btree-inv(rp1, adr2, istore2, istore1, zh, fi7, fi6), btree-inv(rp1, adr2, istore2, istore1, zh, ri, fi3), inv(gstore4 \ ginval3), inv(gstore1), inv(istore2), inv(istore1), ri-inj(fi7), ri-inj(fi6), istore1 ⊑ istore2, ⌊err1⌋ , pstate1.journal? → nx1 ≠ [] ∧ pstate1.lnum = nx1.last ∧ pstate1.lnum ∈ gstore4, pstate1.index? → pstate1.lnum ∈ istore2 ⊦ ⟨if err2 = ESUCCESS then maxino0 := maxino, syn := true, nx0 := nx, ginval1 := ginval, fi0 := fi else ginval0 := ?, syn := ?, gstore0 := ?, gstore := ?, nx := ?, maxino := ?, ginval := ?, maxino0 := ?, nx0 := ?, ginval1 := ?, fi0 := ?, fi := ?⟩ ( fi4 = fi0 ∧ fi5 = fi ∧ ginval2 = ginval0 ∧ gstore2 = gstore0 ∧ gstore3 = gstore ∧ nx2 = nx ∧ maxino1 = maxino ∧ ginval4 = ginval ∧ (syn0 ↔ syn) ∧ maxino2 = maxino0 ∧ nx5 = nx0 ∧ ginval5 = ginval1 ∧ err1 = err2), dups(nx1)

Heuristic symbolic execution applied rule if negative right

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ⌊err⌋ → err ≠ ESUCCESS; (s)