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Proof step 42 from proof nextl-lemma-butlast-nonempty (previous step)

lf(0) = a ', m ≤ n0, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [] ⊦ (m = n0 ⊃ nextl(n0, lf(n0).last);lf(m + 1)) = nextl(m, (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)).last) ∧ (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)) ≠ []

User applied rule case distinction

And obtained the following new goals

Proof step 43:
lf(0) = a ', m ≤ n0, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [] ⊦ (m = n0 ⊃ nextl(n0, lf(n0).last);lf(m + 1)) = nextl(m, (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)).last)
Proof step 48:
lf(0) = a ', m ≤ n0, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [] ⊦ (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)) ≠ []

Rule argument

pos: right 1

Proof step 42 from proof nextl-lemma-butlast-nonempty (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information