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Proof step 6 from proof eq (previous step)

a ' = ins≤(a, a0 ' + y0), (∃ a. a ' = ins≤(a, y0)) → [] = y0 ⊦

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);%empty: ⊦ ins≤(a, x) ≠ [];%ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x;%c: ⊦ a + x = a ' + x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-03: ⊦ a ' = b ' + x ↔ a = b ∧ x = []; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list Used formulas: a0 < a