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Proof step 48 from proof infinite-append-lem (previous step)

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

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 3 flags: discard

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%l-05: ⊦ m + m0 < n → ¬ n + n0 < m; from specification nat-basic2