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Proof step 5 from proof ins-less-first (previous step)

ordered≤(a0 ' + y) ⊦ ins≤(a, a0 ' + y) = a0 ' + ins≤(a, y), a < a0

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: a0 < a

Additional information

Used simplifier rules: ins-n: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x);%ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x;%cons-03: ⊦ ordered≤(a ' + x) → ins≤(a, x) = a ' + x;% ⊦ a + x = a ' + x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ ¬ b < a → (a < b ↔ a ≠ b); (ss)% ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws)%nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem% ⊦ x + y = x + z ↔ y = z; (s) Used formulas: a0 < a