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Proof step 23 from proof inv-inverts (previous step)

nσ[n + 1] = nσ[n] + 1, # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n⌝ , nσ[n] = nσ[m] + 1, m < n, stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[n] ⊦ ((λ m0. max∞(λ n0. nσ[n0] = m0).n) to nσ[n])[nσ[n]] = n, n = 0

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: n + 1 abort: F discard: F computed: F

Additional information

Used simplifier rules: s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic2%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2 Used terms: n + 1