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Proof step 265 from proof inv-format (previous step)

ginval = ginval4, ginval0 = ginval3, ginval1 = ginval3, ginval2 = ginval4, ⌊err0⌋ ⊦ ⟪skip⟫ ( ( err0 = ESUCCESS → syn ∧ (∃ ri, fi. btree-inv(null, adr, istore0, istore, zh, ri, fi)) ∧ inv(gstore0 \ ginval1) ∧ inv(gstore) ∧ elems nx ∩ ginval1 = ∅ ∧ ¬ dups(nx) ∧ dom gstore ∪ elems nx = dom gstore0 ∪ ginval1 ∧ remove-refsize0(gstore) = gstore ∧ inv(istore0) ∧ inv(istore) ∧ istore ⊑ istore0 ∧ remove-refsize0(istore) = istore ∧ (pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore0) ∧ (pstate.index? → pstate.lnum ∈ istore0)) ∧ (err0 = ESUCCESS ∨ ⌊err0⌋ ) ∧ (err0 = ESUCCESS → btree-inv(null, adr, istore0, istore, zh, ∅, ∅) ∧ gstore0 = ∅ ∧ ginval1 = ∅ ∧ maxino = maxino4 ∧ nx = [] ∧ ginval2 = ∅))

Heuristic symbolic execution applied rule skip right

And closed the goal!

Rule argument

Additional information

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