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Proof step 3 from proof ins-one-01 (previous step)

⊦ ins≤(a, b ') = a ' + b ', b < a

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x

Rule argument

spec: <toplevel> name: ins-y seq: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x subst: [b, a, x] → [b, a, []]

Additional information

Used simplifier rules: ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s) Used terms: b, a, []