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Proof step 18 from proof less-leninv (previous step)

nσ[n] = nσ[m] + 1, nσ[n + 1] = nσ[n] + 1, # nσ = ∞, m < n, stutf(nσ), nσ[m] < nσ[n], ∀ m1. m < m1 → nσ[m1] = nσ[n] ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: n + 1 abort: F discard: T 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%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2 Used terms: n + 1