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Proof step 15 from proof infinite-append-lem (previous step)

ii(0) = 0, ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦ m0 ≤ m → (m0 = m + 1 ⊃ ii(m) + # lf(m);ii(m0)) ≤ ii(m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
ii(0) = 0, m0 ≤ m, ii(m) < (m0 = m + 1 ⊃ ii(m) + # lf(m);ii(m0)), ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦ (m0 = m + 1 ⊃ ii(m) + # lf(m);ii(m0)) = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat