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Proof step 19 from proof valid-eraseq-stronger-next-append-none (previous step)

apebs[m1] = mkpeb(aehdr, avhdr, buf, m2, boolvar), apebs[0 + m1] = mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0), m1 + 1 < # apebs, m1 < m, m < # apebs, m1 < # apebs, m1 < m + 1, valid-eraseq-stronger(subarray(apebs, 0, m), q0), ∀ v, m0. (valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m2, boolvar)) ↔ lref = v × m0) ∧ (lref = v × m0 → avhdr.sqn ≠ valid-sqns(v, m0, apebs, m).max), ∀ m, apebs. m < # apebs ∧ valid-eraseq-stronger(subarray(apebs, 0, m), q0) → (valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0 + eq-entry(m, none) ') ↔ ¬ valid-peb(apebs[m])) ⊦ m1 ≠ m ∧ (∀ v, m0. (valid-peb(v, m0, mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0)) ↔ lref = v × m0) ∧ (lref = v × m0 → avhdr0.sqn ≠ (valid-sqns(v, m0, apebs, m) ∪ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)).max)) ∧ valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0 + eq-entry(m, none) ') ↔ ¬ valid-peb(apebs[m]), m1 ∈ q0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
apebs[m1] = mkpeb(aehdr, avhdr, buf, m2, boolvar), m1 < # apebs, m < # apebs, m1 < m, m1 + 1 < # apebs, valid-eraseq-stronger(subarray(apebs, 0, m), q0), ∀ m, apebs. m < # apebs ∧ valid-eraseq-stronger(subarray(apebs, 0, m), q0) → (valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0 + eq-entry(m, none) ') ↔ ¬ valid-peb(apebs[m])), ∀ v, m0. (valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m2, boolvar)) ↔ lref = v × m0) ∧ (lref = v × m0 → avhdr.sqn ≠ valid-sqns(v, m0, apebs, m).max) ⊦ m1 ≠ m ∧ (∀ v, m0. (valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m2, boolvar)) ↔ lref = v × m0) ∧ (lref = v × m0 → avhdr.sqn ≠ (valid-sqns(v, m0, apebs, m) ∪ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)).max)) ∧ valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0 + eq-entry(m, none) ') ↔ ¬ valid-peb(apebs[m]), m1 ∈ q0

Rule argument

Additional information

Used simplifier rules: ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar) = mkpeb(aehdr0, avhdr0, buf0, m0, boolvar0) ↔ aehdr = aehdr0 ∧ avhdr = avhdr0 ∧ buf = buf0 ∧ m = m0 ∧ (boolvar ↔ boolvar0); (s)% ⊦ n < n0 → (n0 < n ↔ false); (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)