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Proof step 5 from proof len (previous step)

s ≠ ∅, ∀ s0. # s0 < # s → true → # set2list(s0) = # s0 ⊦ #(s.min + set2list(s.butmin)) = # s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ s0. # s0 < # s → true → # set2list(s0) = # s0 ⊦ # s = 0, 1 + # set2list(s.butmin) = # s, s = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ # s ≠ 0 → s ≠ ∅; (forward)% ⊦ s ≠ ∅ → # s ≠ 0; (forward)% ⊦ (x + y) + z = x + y + z;% ⊦ # a ' = 0 +1;% ⊦ #(x + y) = # x + # y;% ⊦ a + x = a ' + x;