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Proof step 10 from proof lpt-leb-size-id-idx (previous step)

# lpt1 = # lpt0, lpt0[n] = mklp(m, m0, lpf, m1), LEB_SIZE < m0, n < # lpt0, n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags, ∀ n. n < # lpt0 → lpt1[n].size ≤ LEB_SIZE ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# lpt1 = # lpt0, lpt0[n] = mklp(lpt1[n].ref-size, lpt1[n].size, lpt1[n].flags, m1), n < # lpt0, LEB_SIZE < lpt1[n].size, ∀ n. n < # lpt0 → lpt1[n].size ≤ LEB_SIZE ⊦

Rule argument

Additional information

Used simplifier rules: lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ mklp(n2, n3, lpf, m).ref-size = n2; (s)% ⊦ mklp(n2, n3, lpf, m).size = n3; (s)% ⊦ mklp(n2, n3, lpf, m).flags = lpf; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic