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Proof step 6 from proof abs-istore-cleanup-lpt (previous step)

# lpt = # avol1, # avol0 = # avol1, remove-refsize0(istore)[n] = indexblock(indx, lax, m), lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[n] = mklp(m0, m1, lpf, m2), ∀ n. n < # avol1 ∧ (lpt[n].ref-size ≠ 0 ∨ lpt[n].flags = LP_FREE) → avol0[n] = avol1[n], ∀ n. (n < # avol1 ∧ lpt[n].flags = LP_INDEX_NODES ↔ n ∈ istore) ∧ (n ∈ istore → abs-leb(avol0[n], istore[n], lpt[n]) ∧ lpt[n].ref-size = istore[n].ref-size), ¬(n < # avol1 ∧ lpf = LP_INDEX_NODES ↔ n ∈ remove-refsize0(istore)) ∨ ¬ ( n ∈ remove-refsize0(istore) → abs-leb(avol1[n], indexblock(indx, lax, m0), lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[n]) ∧ m0 = m) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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