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Proof step 13 from proof prefixn-prefixn (previous step)

# y0 = (# y < m ⊃ # y;m), y0 + z0 = y + z ⊦ y = y0, m ≤ # y0 + # z0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# y0 = (# y < m ⊃ # y;m), y0 + z0 = y + z, # y0 + # z0 < m ⊦ y = y0

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)