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Proof step 5 from proof inv-allocate-ind (previous step)

syn ↔ true, ginval = ginval1, ginval0 = ginval2, gstore = gstore1, gstore0 = gstore2, remove-refsize0(gstore1) = gstore1, remove-refsize0(istore0) = istore0, elems nx ∩ ginval2 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval2, inv(gstore2 \ ginval2), inv(gstore1), inv(istore), inv(istore0), istore0 ⊑ istore, pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore2, pstate.index? → pstate.lnum ∈ istore ⊦ ⟪fail_or_success#(; ; err); if err = ESUCCESS then choose n with ¬ n ∈ istore in istore[n] := indexblock([], [], 0), pstate := index(n)⟫ ( inv(istore) ∧ istore0 ⊑ istore ∧ (pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore2) ∧ (pstate.index? → pstate.lnum ∈ istore)), dups(nx)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
remove-refsize0(istore0) = istore0, remove-refsize0(gstore1) = gstore1, elems nx ∩ ginval2 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval2, inv(istore0), inv(istore), inv(gstore1), inv(gstore2 \ ginval2), istore0 ⊑ istore, ¬ dups(nx), pstate.index? → pstate.lnum ∈ istore, pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore2 ⊦ ⟪fail_or_success#(; ; err); if err = ESUCCESS then choose n with ¬ n ∈ istore in istore[n] := indexblock([], [], 0), pstate := index(n)⟫ ( inv(istore) ∧ istore0 ⊑ istore ∧ (pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore2) ∧ (pstate.index? → pstate.lnum ∈ istore))

Proof step 5 from proof inv-allocate-ind (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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