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Proof step 4 from proof id-idx-fwd-1 (previous step)

# lpt1 = # lpt0, lpt1[n] = mklp(m0, m1, lpf1, m2), lpt0[n] = mklp(n0, n1, lpf0, n2), n < # lpt1, ∀ n. n < # lpt1 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags ⊦ n0 = m0 ∧ n1 = m1 ∧ lpf0 = lpf1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# lpt1 = # lpt0, lpt1[n] = mklp(m0, m1, lpf1, m2), lpt0[n] = mklp(n0, n1, lpf0, n2), n < # lpt0, ∀ n. n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags ⊦ n0 = m0 ∧ n1 = m1 ∧ lpf0 = lpf1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic