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Proof step 13 from proof Sep-forall (previous step)

nP0 ⊗_(n + 1, m0) = nP1 ⊗_(n + 1, m0), ∀ n0. ¬ m0 < n0 → nP0(n + n0) = nP1(n + n0) ⊦ nP0(n) * nP1 ⊗_(n + 1, m0) = nP1(n) * nP1 ⊗_(n + 1, m0)

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: 0 abort: F discard: T computed: T

Additional information

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