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

∀ a, b. (a ∈ ins≤(b, x0) ↔ a ∈ x0) ∨ a = b ⊦ a ∈ ins≤(b, a0 ' + x0) ↔ a ∈ a0 ' + x0, a = b

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 6:
a0 < b, ∀ a, b. (a ∈ ins≤(b, x0) ↔ a ∈ x0) ∨ a = b ⊦ a ∈ ins≤(b, a0 ' + x0) ↔ a ∈ a0 ' + x0, a = b

Rule argument

fma: a0 < b

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 → (a ∈ b ' + x ↔ a ∈ x); (s) Used formulas: a0 < b