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Proof step 10 from proof lastn-lastn-01 (previous step)

m0 < n, ∀ n, x. m0 ≤ n → lastn(m0, lastn(n, x)) = lastn(m0, x) ⊦ lastn((m0 + 1)-1, lastn(n, x).butlast) + lastn(n, x).last = lastn(m0, x.butlast) + x.last ', m0 + 1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
m0 < n, ∀ n, x. m0 ≤ n → lastn(m0, lastn(n, x)) = lastn(m0, x) ⊦ lastn(m0, lastn(n, x).butlast) + lastn(n, x).last ' = lastn(m0, x.butlast) + x.last '

Rule argument

Additional information

Used simplifier rules: ⊦ x + a = x + a '; (s)% ⊦ (m + 1)-1 = m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)