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Proof step 5 from proof del1-01-02 (previous step)

ordered≤(a0 ' + x0), ∀ a1, b0. a1 = b0 ∨ ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ⊦ a = b, ins≤(a, (a0 ' + x0) -1l b) = ins≤(a, a0 ' + x0) -1l b

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 6:
a0 < a, ordered≤(a0 ' + x0), ∀ a1, b0. a1 = b0 ∨ ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ⊦ a = b, ins≤(a, (a0 ' + x0) -1l b) = ins≤(a, a0 ' + x0) -1l b
Proof step 11:
ordered≤(a0 ' + x0), ∀ a1, b0. a1 = b0 ∨ ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ⊦ a = b, ins≤(a, (a0 ' + x0) -1l b) = ins≤(a, a0 ' + x0) -1l b, a0 < a

Rule argument

fma: a0 < a

Additional information

Used formulas: a0 < a