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Proof step 5 from proof sizesubl (previous step)

# s0 ≤ # s, s ⊆ s0, ∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1 ⊦ a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# s1 < # s, s ⊆ s1 ++ a ⊦ a ∈ s1, a ∈ s

Rule argument

Additional information

Used simplifier rules: Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)