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Proof step 24 from proof infinite-append-lem (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: m0 abort: F discard: F computed: T

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m0