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Proof step 18 from proof ins-max (previous step)

ordered≤(a0 ' + x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ a ' + a0 ' + x = (a0 ' + x) + a ', x = [], a < (max(x) < a0 ⊃ a0;max(x)), a0 < a, a0 < (max(x) < a0 ⊃ a0;max(x))

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 19:
max(x) < a0, ordered≤(a0 ' + x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ a ' + a0 ' + x = (a0 ' + x) + a ', x = [], a < (max(x) < a0 ⊃ a0;max(x)), a0 < a, a0 < (max(x) < a0 ⊃ a0;max(x))
Proof step 21:
ordered≤(a0 ' + x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ max(x) < a0, a ' + a0 ' + x = (a0 ' + x) + a ', x = [], a < (max(x) < a0 ⊃ a0;max(x)), a0 < a, a0 < (max(x) < a0 ⊃ a0;max(x))

Rule argument

fma: max(x) < a0

Additional information

Used formulas: max(x) < a0