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Proof step 5 from proof sw (previous step)

¬ b < a ⊦ ins≤(a, b ') = ins≤(b, a ')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ a ' + b ' = ins≤(b, a '), b < a

Rule argument

Additional information

Used simplifier rules: ins-one-01: ⊦ ¬ b < a → ins≤(a, b ') = a ' + b ';