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Proof step 4 from proof app-03 (previous step)

¬ a0 ∈ y0 → y0 = [], ∀ a1. #oc(a1, a ') + #oc(a1, y0) = 0 ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a abort: F discard: T computed: F

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1;% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: a