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Proof step 176 from proof inv-format (previous step)

err = err0, ginval = ∅, ginval0 = ∅, gstore = ∅, gstore0 = ∅, istore = ∅, istore0 = (∅ )[n0, indexblock (indexnode(mkbrarray(BRANCH_SIZE), true, 0) ', @(maxino2, indexnode(mkbrarray(BRANCH_SIZE), true, 0).flashsize) ', indexnode(mkbrarray(BRANCH_SIZE), true, 0).flashsize) ], maxino = maxino4, nx = [], inv((∅)[n0, indexblock([], [], 0)]), inv((∅)[n0, indexblock(indexnode(mkbrarray(BRANCH_SIZE), true, 0) ', @(maxino2, indexnode(mkbrarray(BRANCH_SIZE), true, 0).flashsize) ', 0)]), inv ((∅ )[n0, indexblock (indexnode(mkbrarray(BRANCH_SIZE), true, 0) ', @(maxino2, indexnode(mkbrarray(BRANCH_SIZE), true, 0).flashsize) ', indexnode(mkbrarray(BRANCH_SIZE), true, 0).flashsize) ]), indexnode(mkbrarray(BRANCH_SIZE), true, 0).flashsize ≤ LEB_SIZE, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪ginval1 := ?⟫ ( ( err = ESUCCESS → (∃ ri, fi. btree-inv(null, adr, istore0, istore, zh, ri, fi)) ∧ inv(gstore0 \ ginval0) ∧ inv(gstore) ∧ elems nx ∩ ginval0 = ∅ ∧ ¬ dups(nx) ∧ dom gstore ∪ elems nx = dom gstore0 ∪ ginval0 ∧ remove-refsize0(gstore) = gstore ∧ inv(istore0) ∧ inv(istore) ∧ istore ⊑ istore0 ∧ remove-refsize0(istore) = istore ∧ (readonly.journal? → nx ≠ [] ∧ readonly.lnum = nx.last ∧ readonly.lnum ∈ gstore0) ∧ (readonly.index? → readonly.lnum ∈ istore0)) ∧ (err = ESUCCESS ∨ ⌊err⌋ ) ∧ (err = ESUCCESS → btree-inv(null, adr, istore0, istore, zh, ∅, ∅) ∧ gstore0 = ∅ ∧ ginval0 = ∅ ∧ maxino = maxino4 ∧ nx = [] ∧ ginval = ∅)), err = ESUCCESS

Proof step 176 from proof inv-format (previous step)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

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