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Proof step 25 from proof diff-sub-size (previous step)

# s0 = #(s0 \ s) + # s, a ∈ s0, a0 ∈ s0, # s ≤ # s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #(s0 \ s) - 1 = # s0 - # s + 1, a ∈ s

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 26:
n ≤ m ⊦ n0 = m - n ↔ m = n0 + n
Proof step 27:
# s0 = #(s0 \ s) + # s, a0 ∈ s0, a ∈ s0, # s ≤ # s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #(s0 \ s) - 1 = # s0 - # s + 1, a ∈ s, # s + 1 ≤ # s0

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [n, m, n0] → [# s + 1, # s0, #(s0 \ s) - 1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)% ⊦ ¬ a ∈ ∅; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s) Used terms: # s + 1, # s0, #(s0 \ s) - 1