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Proof step 56 from proof nextl-nextj-lemma-butlast (previous step)

lf(0) = a ', lf(n + 1) = nextl(jj(n), lf(n).last), jj(n + 1) = nextj(jj(n), lf(n).last), n ≤ m0 ⊦ n = m0 + 1, lf(n + 1) = nextl((n = m0 + 2 ⊃ nextj(jj(m0 + 1), lf(m0 + 1).last);jj(n)), (n = m0 + 2 ⊃ nextl(jj(m0 + 1), lf(m0 + 1).last);lf(n)).last) ∧ jj(n + 1) = nextj((n = m0 + 2 ⊃ nextj(jj(m0 + 1), lf(m0 + 1).last);jj(n)), (n = m0 + 2 ⊃ nextl(jj(m0 + 1), lf(m0 + 1).last);lf(n)).last)

User applied rule cut formula

And closed the goal!

Rule argument

fma: n = m0 + 2

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat Used formulas: n = m0 + 2