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Proof step 3 from proof in-04 (previous step)

∀ a. a ∈ ins≤(a, x0) ⊦ a ∈ ins≤(a, a0 ' + x0)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a0 < a, ∀ a. a ∈ ins≤(a, x0) ⊦ a ∈ ins≤(a, a0 ' + x0)

Rule argument

fma: a0 < a

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x;% ⊦ a + x = a ' + x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' + x ↔ ¬(a ≠ b ∧ ¬ a ∈ x); (ss) Used formulas: a0 < a