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Proof step 50 from proof eq-leninv (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
# nσ = ∞, m + 1 = nσ[n], stutf(nσ), ∀ m0. n < m0 → nσ[m0] = nσ[n] ⊦ ⌜m⌝ + ⌜1⌝ = leninv(nσ)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)