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Proof step 7 from proof Sep-range (previous step)

∀ nP, n. nP ⊗_(n, m0) = (λ n0. ⌜n0 < n + m0⌝ → nP(n0))⊗_(n, m0) ⊦ nP(n) * nP ⊗_(n + 1, m0) = (⌜¬ n + m0 < n⌝ → nP(n)) * (λ n0. ⌜¬ n + m0 < n0⌝ → nP(n0))⊗_(n + 1, m0)

User applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: nP, n + 1 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ¬ m + n < m; (s)% ⊦ ⌜true⌝ = true; (s)% ⊦ (true → P) = P; (s) Used terms: nP, n + 1