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Proof step 5 from proof id-idx-fwd-0 (previous step)

# 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

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ mklp(m, m0, lpf, m1).flags = lpf; (s)% ⊦ mklp(m, m0, lpf, m1).size = m0; (s)% ⊦ mklp(m, m0, lpf, m1).ref-size = m; (s) Used terms: n