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Proof step 3 from proof lastn-lastn-01 (previous step)

∀ n, x. m0 ≤ n → lastn(m0, lastn(n, x)) = lastn(m0, x), m0 +1 ≤ n ⊦ lastn(m0 +1, lastn(n, x)) = lastn(m0 +1, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m0 < n, ∀ n, x. m0 ≤ n → lastn(m0, lastn(n, x)) = lastn(m0, x) ⊦ lastn(m0 + 1, lastn(n, x)) = lastn(m0 + 1, x), n = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2