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Proof step 19 from proof prefixn-prefixn (previous step)

# (m ≤ # x ⊃ firstn(m, x);x) < n ⊦ x = ((n < m ⊃ n;m) ≤ # x ⊃ firstn((n < m ⊃ n;m), x);x), m ≤ # x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
# x < m, # x < n ⊦ x = ((n < m ⊃ n;m) ≤ # x ⊃ firstn(n, x);x)

Rule argument

Additional information

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