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Proof step 14 from proof prefixn-prefixn (previous step)

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0); (ws)%a: ⊦ (x + y) + z = x + y + z; from specification list