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Proof step 1 from proof size-gtleq (conclusion)

⊦ b < a ∧ ¬ b ∈ s → #(s ++ a ++ b) = #(s ++ a) + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ ¬ a ∈ s → #(s ++ a) = # s + 1

Rule argument

spec: set-basic inst: name: Size-insert seq: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1 subst: [a, s] → [b, s ++ a] flags: rewrite

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s; from specification set-basic%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%in-03: ⊦ ¬ a ∈ s → (a ∈ s ++ b ↔ a = b); from specification set-basic%lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem%irreflexivity: ⊦ ¬ a < a; from specification oelem Used terms: b, s ++ a