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Proof step 15 from proof prefixn-prefixn (previous step)

(m ≤ # x ⊃ firstn(m, x);x) = y + z ⊦ y = x, (# y < m ⊃ # y;m) ≤ # x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
(m ≤ # x ⊃ firstn(m, x);x) = y + z, # x < (# y < m ⊃ # y;m) ⊦ y = x

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)