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Proof step 7 from proof union-sizele (previous step)

#(s ∪ s0) ≤ # s ⊦ #(s ∪ s0) < # s, #(s ∪ s0) = # s, ∀ a. a ∈ s0 → a ∈ s

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ m ≠ n → (¬ m < n ↔ n < m)

Rule argument

spec: nat-basic1 inst: name: neq-01 seq: ⊦ m ≠ n → (¬ m < n ↔ n < m) subst: [n, m] → [# s, #(s ∪ s0)] flags: rewrite

Additional information

Used simplifier rules: union-sizeeq: ⊦ #(s ∪ s0) = # s ↔ s0 ⊆ s;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic%In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used terms: # s, #(s ∪ s0)