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Proof step 5 from proof id-idx-write-both (previous step)

# lpt1 = # lpt0, ∀ n. n < # lpt1 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags ⊦ id-idx(lpt0[n, mklp(n0, n1, lpf, n2)], lpt1[n, mklp(n0, n1, lpf, m2)])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# lpt1 = # lpt0, ∀ n. n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags ⊦ id-idx(lpt0[n, mklp(n0, n1, lpf, n2)], lpt1[n, mklp(n0, n1, lpf, m2)])

Proof step 5 from proof id-idx-write-both (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information