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Proof step 6 from proof sep-forall-reduce-true (previous step)

((nP ⊗_(n, m0 + 1) * P0))(zh), m0 < m0 + 1, ∀ n0, P. ¬ m0 < n0 ∧ ((nP(n + n0) * P))(zh) → ((nP2(n + n0) * true * P))(zh), ∀ nP, n, P0, zh, nP2. ((nP ⊗_(n, m0) * P0))(zh) ∧ (∀ n0, P. n0 < m0 ∧ ((nP(n + n0) * P))(zh) → ((nP2(n + n0) * true * P))(zh)) → ((nP2 ⊗_(n, m0) * true * P0))(zh) ⊦ (((nP2 ⊗_(n, m0) * nP2(n + m0) * nP2 ⊗_(n + m0 + 1, m0 + 1 - m0 - 1)) * true * P0))(zh)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
((nP ⊗_(n, m0 + 1) * P0))(zh), ∀ nP, n, P0, zh, nP2. ((nP ⊗_(n, m0) * P0))(zh) ∧ (∀ n0, P. n0 < m0 ∧ ((nP(n + n0) * P))(zh) → ((nP2(n + n0) * true * P))(zh)) → ((nP2 ⊗_(n, m0) * true * P0))(zh), ∀ n0, P. ¬ m0 < n0 ∧ ((nP(n + n0) * P))(zh) → ((nP2(n + n0) * true * P))(zh) ⊦ ((nP2 ⊗_(n, m0) * nP2(n + m0) * true * P0))(zh)

Rule argument

Additional information

Used simplifier rules: sep-com: ⊦ P * P2 = P2 * P; from specification heap-sep[zheap]%sep-ass: ⊦ (P * P3) * P4 = P * P3 * P4; from specification heap-sep[zheap]% ⊦ emp * P0 = P0; (s)% ⊦ nP0 ⊗_(n, 0) = emp; (s)% ⊦ m + n - m = n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)