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Proof step 52 from proof infinite-append-lem (previous step)

ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), ii(m1 + 1) = ii(m1) + # lf(m1), (nf from ii(m)) to # lf(m) = lf(m), (nf from ii(m1)) to # lf(m1) = lf(m1), n1 + ii(m1 + 1) < ii(m1), m < m1, n < # lf(m), n1 < # lf(m1 + 1), ii(m + 1) ≤ ii(m1), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m2. ¬ m1 < m2 → ii(m2 + 1) = ii(m2) + # lf(m2) ∧ (nf from ii(m2)) to # lf(m2) = lf(m2) ∧ (∀ m0. m0 ≤ m2 → ii(m0) ≤ ii(m2)), ∀ m0. ¬ m1 < m0 → ii(m0) ≤ ii(m1) ⊦ m1 + 1 = m, lf(m1 + 1)[n1] = lf(m)[n], lf(m) = [], lf(m1 + 1) = [], m = 2 + m1, m1 = m

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 3 flags: discard

Additional information

Used simplifier rules: ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic